Contoh 5 Soal dan Pembahasan Materi Hubungan Dua Lingkaran

$\begin{array}{l}\\ 21.& \text{Titik Kuasa dari lingkaran-lingkaran}\\ & \text{berikut}\\ & L_1\equiv x^2+y^2+x+y-14=0\\ & L_2\equiv x^2+y^2=13\\ & L_3\equiv x^2+y^2+3x-2y-26=0\text{adalah}....\\ & \text{a}.(3,-2)\\ & \text{b}.(2,-3)\\ & \text{c}.(-3,2)\\ & \text{d}.(-2,3)\\ & \text{e}.(3,2)\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan eliminasi, kita mendapatkan}\\ & \begin{array}{rl}& \begin{array}{lrll}(L_1)& x^2+y^2+x+y& =& 14\\ (L_2)& x^2+y^2& =& 13& -\\ & x+y& =& 1& ....(1)\end{array}\\ & \text{dan}\\ & \begin{array}{lrll}(L_3)& x^2+y^2+3x-2y& =& 26\\ (L_2)& x^2+y^2& =& 13& -\\ & 3x-2y& =& 13& ....(2)\end{array}\\ & \text{Selanjutnya kita eliminasi}(1)\mathrm{\&}(2)\\ & \text{dan hasilnya adalah}:\\ & \begin{array}{rrlr}(2)& 3x-2y& =& 13\\ (1)& 3x+3y& =& 3& -(\times 3)\\ & -5y& =& 10& \\ & y& =& -2& \Rightarrow x=3\end{array}\\ & \text{Jadi, titik kuasa ketiganya}:(3,-2)\end{array}\\ & \mathbf{\text{Sebagai ilustrasi perhatikan gambar berikut}}\end{array}$

$\begin{array}{l}\\ 22.& \text{Titik-titik potong dari persekutuan dua}\\ & \text{lingkaran}{L}_1\equiv (x-2{)}^2+y^2=10\text{dan}\\ & L_2\equiv x^2+(y-2{)}^2=10\text{adalah}....\\ & \text{a}.(3,3)\text{dan}(1,1)\\ & \text{b}.(3,3)\text{dan}(-1,-1)\\ & \text{c}.(3,-3)\text{dan}(1,1)\\ & \text{d}.(-3,3)\text{dan}(1,1)\\ & \text{e}.(-3,-3)\text{dan}(-1,-1)\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Alternatif 1}\\ & \text{Dengan substitusi opsi pilihan jawaban}\\ & \text{maka akan ketemu jawabannya langsung}\\ & \text{Alternatif 2}\\ & \text{Dengan eliminasi dan ilustrasi gambar}\\ & \begin{array}{lrll}(L_1)& (x-2{)}^2+y^2& =& 10\\ (L_2)& x^2+(y-2{)}^2& =& 10& \\ & \text{menjadi}\\ (L_1)& x^2+y^2-4x& =& 6\\ (L_2)& x^2+y^2-4y& =& 6& -\\ & -4x+4y& =& 0& \\ & \text{maka hasilnya}\\ & y& =& x\end{array}\\ & \text{Jelas opsi jawaban c, d salah}\\ & \text{karena}y=x,\\ & \mathbf{\text{Dengan bantuan ilustrasi, pilihan jawaban}}\\ & \mathbf{\text{akan tampak dengan jelas}}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Persamaan tali busur persekutuan dua}\\ & \text{lingkaran}{L}_1\equiv (x-3{)}^2+y^2=16\text{dan}\\ & L_2\equiv x^2+(y-3{)}^2=16\text{adalah}....\\ & \text{a}.y=-2x\\ & \text{b}.y=-x\\ & \text{c}.y=x\\ & \text{d}.y=2x\\ & \text{e}.y={\displaystyle \frac{1}{2}x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan eliminasi, kita mendapatkan}\\ & \begin{array}{lrll}(L_1)& (x-3{)}^2+y^2& =& 16\\ (L_2)& x^2+(y-3{)}^2& =& 16& \\ & \text{menjadi}\\ (L_1)& x^2+y^2-6x& =& 9\\ (L_2)& x^2+y^2-6y& =& 9& -\\ & -6x+6y& =& 0& \\ & \text{maka hasilnya}\\ & y& =& x\end{array}\\ & \mathbf{\text{Sebagai ilustrasi perhatikan gambar berikut}}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Banyaknya garis singgung persekutuan}\\ & \text{lingkaran-lingkaran}{x}^2+y^2+2x-6y+9=0\\ & \text{dan}{x}^2+y^2+8x-6y+9=0\text{adalah}....\\ & \text{a}.0\\ & \text{b}.1\\ & \text{c}.2\\ & \text{d}.3\\ & \text{e}.4\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{|ll|}\hline \text{Lingakaran}& \text{Pusat/r}\\ L_1\equiv x^2+y^2+2x-6y+9=0& \{\begin{array}{ll}{P}_1& =(-1,3)\\ r_1& =1\end{array}\\ L_2\equiv x^2+y^2+8x-6y+9=0& \{\begin{array}{ll}{P}_2& =(-4,3)\\ r_2& =4\end{array}\\ \hline\end{array}\\ & \text{Perhatikan pula bahwa}{r}_2-r_1=4-1=3\\ & \begin{array}{rl}& \text{Karena}{P}_1{P}_2=r_2-r_1,\text{hal ini berarti lingkaran}\\ & L_1\text{bersinggungan di dalam dengan lingkaran}{L}_2\\ & \text{Sehingga kedua lingkaran ini hanya akan }\\ & \text{memiliki}\text{satu}\text{garis singgung persekutuan}\end{array}\\ & \mathbf{\text{Sebagai ilustrasi perhatikan gambar berikut}}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Persamaan lingkaran dengan jari-jari}5\\ & \text{dan menyinggung lingkaran lain}\\ & x^2+y^2-2x-4y-20=0\text{di titik}\\ & (5,5)\text{adalah}....\\ & \text{a}.x^2+y^2-2x-4y-120=0\\ & \text{b}.x^2+y^2-2x-4y-120=0\\ & \text{c}.x^2+y^2-2x-4y-120=0\\ & \text{d}.x^2+y^2-2x-4y-120=0\\ & \text{e}.x^2+y^2-2x-4y-120=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahi bahwa}\\ & \begin{array}{rl}& \begin{array}{rrll}(L_1)& (x-a{)}^2+(y-b{)}^2& =& 5^2\\ (L_2)& x^2+y^2-2x-4y& =& 20& \end{array}\\ & \text{Titik singgung dua lingkaran}\\ & \text{di titik}(5,5),\text{artinya}\\ & \left(\begin{array}{c}5\\ 5\end{array}\right)={\displaystyle \frac{\left(\begin{array}{c}a\\ b\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)}{2}}\\ & \iff \left(\begin{array}{c}10\\ 10\end{array}\right)=\left(\begin{array}{c}a\\ b\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & \iff \left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}10-1\\ 10-2\end{array}\right)=\left(\begin{array}{c}9\\ 8\end{array}\right)\end{array}\\ & \begin{array}{rl}& \text{maka persamaan lingkarannya adalah}:\\ & \iff (x-9{)}^2+(y-8{)}^2=5^2\\ & \iff x^2+y^2-18x-16y+120=0\end{array}\\ & \mathbf{\text{Berikut ilustrasi gambarnya}}\end{array}$.

Operasi Vektor di Ruang (Lanjutan Materi Vektor di Ruang)

Sebelumnya silahkan lihat di sini

 $\text{A. Operasi Vektor Dalam Ruang}$

Operasi vektor pada dimensi tiga kurang lebih sama dengan operasi pada vektor berdimensi dua.

$\text{A. Penjumlahan dan Pengurangan}$.

$\begin{array}{rl}\text{Jika}& \text{diketahui sebagai misal}\\ \overline{u}& =a\overline{i}+b\overline{j}+c\overline{k}\text{dan}\\ \overline{v}& =p\overline{i}+q\overline{j}+r\overline{k}\\ \text{mak}& \text{a}\\ \mathbf{\text{Pen}}& \mathbf{\text{jumlahan dua vektor di atas adalah}}\\ \overline{u}+\overline{v}& =(a+p)\overline{i}+(b+q)\overline{j}+(c+r)\overline{k}\\ \mathbf{\text{dem}}& \mathbf{\text{ikian juga untuk pengurangan}}\\ \overline{u}-\overline{v}& =(a-p)\overline{i}+(b-q)\overline{j}+(c-r)\overline{k}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika diketahui}\overline{a}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & \text{Tentukanlah hasil dari}\\ & \text{a}.\overline{a}+\overline{b}\\ & \text{b}.\overline{a}-\overline{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \overline{a}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right),\\ & \text{maka}\\ & \overline{a}+\overline{b}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)+\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}1+8\\ 3+(-2)\\ 7+0\end{array}\right)=\left(\begin{array}{c}9\\ 1\\ 7\end{array}\right)\\ & \text{Dan untuk}\overline{a}-\overline{b}\text{adalah}:\\ & \overline{a}-\overline{b}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)-\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}1-8\\ 3-(-2)\\ 7-0\end{array}\right)=\left(\begin{array}{c}-7\\ 5\\ 7\end{array}\right)\end{array}\end{array}$.

$\text{B. 1. Perkalian Skalar dengan Vektor}$.

Misalkan suatu skalar   $m$  dan suatu vektor  $\overline{u}=a\overline{i}+b\overline{j}+c\overline{k}$, maka perkalian $m$  dengan vektor  $\overline{u}$ tersebut adalah  $\overline{u}=ma\overline{i}+mb\overline{j}+mc\overline{k}$.

$\text{CONTOH SOAL}$.

$\begin{array}{ll}2.& \text{Jika}\overline{a}=\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right),\text{tentukanlah nilai}\\ & \text{dari}2\overline{a}\text{dan}-3\overline{a}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}2\overline{a}& =2\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right)=\left(\begin{array}{c}4044\\ 4042\\ 4040\end{array}\right),\text{dan}\\ -3\overline{a}& =-3\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right)=\left(\begin{array}{c}-6066\\ -6063\\ -6060\end{array}\right)\end{array}\end{array}$

$\text{F. 2. Perkalian Skalar Dua Vektor}$.

Hasil dari perkalian skalar dua vektor $\overline{a}$  dan  $\overline{b}$ adalah :  $\overline{a}\bullet \overline{b}$.

Dengan

$\overline{a}\bullet \overline{b}=\left|\overline{a}\right|\left|\overline{b}\right|\cos \theta$.  sehingga

$\begin{array}{rl}& \text{Tanda dari hasil skalar ini adalah}\\ & \begin{array}{|lll|}\hline \mathbf{\text{Besar sudut}}\theta & \mathbf{\text{Tanda}}& \text{Bentuk}\\ 0^{\circ }\le \theta <{90}^{\circ }& \text{Positif}& \text{Lancip}\\ \theta ={90}^{\circ }& \text{Nol}& \text{Siku-siku}\\ {90}^{\circ }<\theta \le {180}^{\circ }& \text{Negatif}& \text{Tumpul}\\ \hline\end{array}\\ & \text{Untuk}\theta \text{berupa sudut istimewa}:\\ & \begin{array}{|ccccccc|}\hline \theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }\\ \cos \theta & 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1\\ \hline\end{array}\end{array}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

$\cos \theta ={\displaystyle \frac{\overline{a}\bullet \overline{b}}{\left|\overline{a}\right|\left|\overline{b}\right|}}$.

Sebagai ilustrasinya perhatikanlah gambar berikut

Selain hasil di atas ada cara lain menyelesaikan perkalian skalar dua vektor, yaitu:

$\begin{array}{rl}\text{Jika}& \text{diketahui sebagai misal}\\ \overline{u}& =a\overline{i}+b\overline{j}+c\overline{k}\text{dan}\\ \overline{v}& =p\overline{i}+q\overline{j}+r\overline{k}\\ \text{mak}& \text{a}\\ \mathbf{\text{Per}}& \mathbf{\text{kalian skalar dua vektor adalah}}:\\ \overline{u}\bullet & \overline{v}=(a\overline{i}+b\overline{j}+c\overline{k})(p\overline{i}+q\overline{j}+r\overline{k})\\ & =ap.\overline{i}\bullet \overline{i}+aq.\overline{i}\bullet \overline{j}+ar.\overline{i}\bullet \overline{k}\\ & +bp.\overline{j}\bullet \overline{i}+bq.\overline{j}\bullet \overline{j}+br.\overline{j}\bullet \overline{k}\\ & +cp.\overline{k}\bullet \overline{i}+cq.\overline{k}\bullet \overline{j}+cr.\overline{k}\bullet \overline{k}\\ & =ap+0+0+0+bq+0+0+0+cr\\ & =ap+bq+cr\end{array}$

$\begin{array}{rl}& \text{Sebagai penjelasannya adalah}:\\ & ▹\overline{i}\bullet \overline{i}=\left|\overline{i}\right|\left|\overline{i}\right|\cos 0^{\circ }=1.1.1=1\\ & ▹\overline{i}\bullet \overline{j}=\left|\overline{i}\right|\left|\overline{j}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{i}\bullet \overline{k}=\left|\overline{i}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{j}\bullet \overline{i}=\overline{i}\bullet \overline{j}=0\\ & ▹\overline{j}\bullet \overline{j}=\left|\overline{j}\right|\left|\overline{j}\right|\cos 0^{\circ }=1.1.1=1\\ & ▹\overline{j}\bullet \overline{k}=\left|\overline{j}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{k}\bullet \overline{i}=\overline{i}\bullet \overline{k}=0\\ & ▹\overline{k}\bullet \overline{j}=\overline{j}\bullet \overline{k}=0\\ & ▹\overline{k}\bullet \overline{k}=\left|\overline{k}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.1=1\end{array}$

$\begin{array}{rl}& \text{Atau jika ditabelkan nilainya}\\ & \begin{array}{|cccc|}\hline \overline{u}\bullet \overline{v}& p\overline{i}& q\overline{j}& r\overline{k}\\ a\overline{k}& ap& 0& 0\\ b\overline{j}& 0& bq& 0\\ c\overline{k}& 0& 0& cr\\ \hline\end{array}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{ll}3.& \text{Jika}\overrightarrow{a}=\left(\begin{array}{c}1\\ 2\\ 4\end{array}\right),\text{dan}\overrightarrow{b}=\left(\begin{array}{c}5\\ 4\\ 0\end{array}\right)\\ & \text{tentukanlah nilai}\text{dari}\overrightarrow{a}\bullet \overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =1.5+2.4+4.0=5+8+0=13\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Jika diketahui}\overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k},\\ & \text{dan}\overrightarrow{b}=3\overrightarrow{i}-4\overrightarrow{j}+m\overrightarrow{k}\text{serta}\\ & \text{nilai}\overrightarrow{a}\bullet \overrightarrow{b}=-4,\text{maka tentukan}\\ & \text{nilai}m\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Diketahui bahwa}\\ & ▹\overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right),\text{dan}\\ & ▹\overrightarrow{b}=3\overrightarrow{i}-4\overrightarrow{j}+m\overrightarrow{k}=\left(\begin{array}{c}3\\ -4\\ m\end{array}\right)\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =1.3+3.(-4)+3.m\\ -4& =3+8+3m\\ -3m& =11+4\\ m& =-{\displaystyle \frac{15}{3}}\\ & =-5\end{array}\end{array}$

$\begin{array}{ll}5.& \text{Diketahui}\left|\overrightarrow{a}\right|=10,\left|\overrightarrow{b}\right|=6.\\ & \text{Jika}\overrightarrow{a}\text{dan}\overrightarrow{b}\text{membentuk sudut}\\ & {60}^{\circ }.\text{Tentukanlah nilai}\overrightarrow{a}\bullet \overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ & =10.6.\cos {60}^{\circ }\\ & =60.\left({\displaystyle \frac{1}{2}}\right)\\ & =30\\ \text{Jadi}& \text{hasil kali skalarnya adalah 30}\end{array}\end{array}$.

Contoh Soal 3 Distribusi Normal

 $\begin{array}{l}\\ 11.& \text{Tersiar kabar bahwa harga beras di pasar di}\\ & \text{wilayah B adalah}\text{Rp}8.000,00/\text{Kg dengan}\\ & \text{simpangan baku}\text{Rp}1.500,00.\text{Berdasar kabar}\\ & \text{tersebut dilakukan penelitian dengan mengambil}\\ & \text{sampel secara acak sebanyak}60\text{kios yang dan}\\ & \text{diperoleh rata-rata harga beras}\text{Rp}7.800,00/\text{Kg}\\ & \text{Jika penghitungan menggunakan tingkat}\\ & \text{signifikansi}5\mathrm{\%},\text{maka kesimpulan berikut }\\ & \text{yang tepat adalah}....\\ & \text{a}.\text{harga beras di pasar lebih dari}\text{Rp}7.800,00/\text{Kg}\\ & \text{b}.\text{harga beras di pasar lebih dari}\text{Rp}8.000,00/\text{Kg}\\ & \text{c}.\text{harga beras di pasar kurang dari}\text{Rp}0.000,00/\text{Kg}\\ & \text{d}.\text{harga beras di pasar}\text{Rp}7.800,00/\text{Kg}\\ & \text{e}.\text{harga beras di pasar}\text{Rp}8.000,00/\text{Kg}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{―}{\text{Hipotesis}}\\ & \text{Rata-rata harga beras dipasar}\text{Rp}8.000,00/\text{Kg}\\ & {\text{H}}_0:\mu =8.000\\ & {\text{H}}_0:\mu \ne 8.000\\ & \underset{―}{\text{Daerah Kritis}}\\ & \text{Taraf nyata yang dipilih adalah}=\alpha =0,05=5\mathrm{\%}\\ & {\displaystyle \frac{\alpha }{2}=2,5\mathrm{\%}=0,025}\\ & {\text{z}}_{0,025}=1,96\\ & \text{maka daerah kritis/penolakannya adalah}\\ & \text{z}<-1,96\text{atau}\text{z}>1,96\end{array}\end{array}$.

$.\begin{array}{rl}& \underset{―}{\text{Nilai Satistik Uji}}\\ & \text{Dihitung dengan rumus}:\text{z}={\displaystyle \frac{\bar{\text{x}}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}}}\\ & \text{z}={\displaystyle \frac{7.800-8.000}{{\displaystyle \frac{1.500}{\sqrt{60}}}}=-{\displaystyle \frac{200\sqrt{60}}{1.500}=-1,03}}\\ & \underset{―}{\text{Keputusan Uji}}\\ & \text{Karena nilai}-1,96<\text{z}<1,96,\\ & \text{maka}{\text{H}}_0\text{diterima}\\ & \underset{―}{\text{Kesimpulan}}\\ & \text{Rata-rata harga beras dipasar}\text{Rp}8.000,00/\text{Kg}\end{array}$ .

Contoh Soal 2 Distribusi Normal

$\begin{array}{l}\\ 6.& \text{Luas daerah yang diarsir di bawah}\\ & \text{kurva normal baku berikut adalah}....\\ & \text{a}.0,4861\\ & \text{b}.0,4878\\ & \text{c}.0,4881\\ & \text{d}.0,4938\\ & \text{e}.0,4946\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& & & & & & \downarrow & & & & \\ ⋮& & & & & & & & & & \\ 2,2& & & & & & 0,4878& & & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga nilai}\text{z}=2,25\text{luasnya}=0,4878\end{array}$.

$\begin{array}{l}\\ 7.& \text{Luas daerah yang diarsir di bawah}\\ & \text{kurva normal baku berikut adalah}....\\ & \text{a}.0,1138\\ & \text{b}.0,3810\\ & \text{c}.0,3862\\ & \text{d}.0,4948\\ & \text{e}.0,5000\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& & & & & & & \downarrow & & \downarrow & \\ ⋮& & & & & & & & & & \\ 1,1& & & & & & & \downarrow & & 0,3810& \\ ⋮& & & & & & & & & & \\ 2,5& & & & & & & 0,4948& & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga nilai}\text{z}=1,18\text{luasnya}=0,3810\\ & \text{Dan nilai}\text{z}=2,56\text{luasnya}=0,4948\\ & \text{maka luas arsiran}\\ & =P(1,18<Z<2,56)\\ & =P(0<Z<2,56)-P(0<Z<1,18)\\ & =0,4948-0,3810\\ & =0,1138\end{array}$.

$\begin{array}{l}\\ 8.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah adalah 0,9332, maka nilai}\text{z}=....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.1,05\\ & \text{b}.1,06\\ & \text{c}.1,16\\ & \text{d}.1,50\\ & \text{e}.1,60\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran adalah}\\ & =P(Z<z)=0,9332\\ & =0,5+0,4332\\ & =0,5+P(0<Z<z)\\ & \text{lihat/konfirmasi ke tabel}\\ & \text{z}=1,50\\ & \begin{array}{rl}& \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& \downarrow & & & & & & & & & \\ ⋮& & & & & & & & & & \\ 1,5& 0,4332& & & & & & & & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga luasnya}=0,4878,\text{batas z}=1,50\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Pada suatu kelas seorang guru matematika}\\ & \text{menyatakan bahwa nilai ulangan mapel }\\ & \text{yang diampunya tidak kurang dari}68\\ & \text{Untuk menentukan uji tersebut, maka guru}\\ & \text{yang bersangkutan memilih 10 anak secara}\\ & \text{acak untuk ditanyai nilai hasil ulangannya}\\ & \text{Hipotesis}{\text{H}}_0\text{dan}{\text{H}}_1\text{yang tepat dari kondisi}\\ & \text{kondisi di atas adalah}....\\ & \text{a}.\begin{array}{c}{\text{H}}_0:\mu =68\\ {\text{H}}_1:\mu \ne 68\end{array}\\ & \text{b}.\begin{array}{c}{\text{H}}_0:\mu =68\\ {\text{H}}_1:\mu >68\end{array}\\ & \text{c}.\begin{array}{c}{\text{H}}_0:\mu \ge 68\\ {\text{H}}_1:\mu <68\end{array}\\ & \text{d}.\begin{array}{c}{\text{H}}_0:\mu \le 68\\ {\text{H}}_1:\mu >q68\end{array}\\ & \text{e}.\begin{array}{c}{\text{H}}_0:\mu >68\\ {\text{H}}_1:\mu \ne 68\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \text{Cukup jelas}\\ & \text{Dan ini contoh uji satu pihak, yaitu kiri}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Seorang petugas}customerservice\text{menyatakan}\\ & \text{bahwa di dealer A dapat mensevis rata-rata }\\ & \text{sebanyak}75\text{unit sepeda motor perhari}.\text{Untuk}\\ & \text{mengecek kebenaran pernyataan di atas diambil}\\ & \text{sampel beberapa hari secara random sebanyak }\\ & \text{20 hari. Dari penelitian ini diperoleh rata-rata}\\ & \text{78 unit dengan simpangan bakunya 5. Hasil}\\ & \text{perhitungan}{\text{z}}_0-\text{nya adalah}....\\ & \text{a}.2,35\\ & \text{b}.2,43\\ & \text{c}.2,55\\ & \text{d}.2,68\\ & \text{e}.2,75\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui data dianggap berdistribusi}\\ & \text{normal baku}N(0,1)\text{dengan}\\ & \text{Rata-rata sampel}=\bar{\text{x}}=78\text{unit sepeda motor}\\ & \text{Rata-rata populasinya yang diuji}={\mu }_0=75\\ & \text{Simpangan bakunya}=\sigma =5\text{unit}\\ & \text{dengan banyak data sampel}=n=20\text{hari}\\ & \text{Penghitungan nilainya}z-\text{nya}\\ & ={\displaystyle \frac{\bar{\text{x}}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}}}\\ & ={\displaystyle \frac{78-75}{{\displaystyle \frac{5}{\sqrt{20}}}}={\displaystyle \frac{3}{{\displaystyle \frac{5}{2\sqrt{5}}}}}}\\ & ={\displaystyle \frac{6}{\sqrt{5}}=\frac{6}{2,236}=2,683}\end{array}\end{array}$.

Contoh Soal 1 Distribusi Normal

 $\begin{array}{l}\\ 1.& \text{Fungsi distribusi normal variabel acak X}\\ & \text{dengan}\mu =8\text{dan}\sigma =2\text{adalah}....\\ & \text{a}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{2}}}}}\\ & \text{b}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{4}}}}}\\ & \text{c}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{2}}}}}\\ & \text{d}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{8\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{4}}}}}\\ & \text{e}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{8\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{8}}}}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle f(x)}& ={\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}}},\text{dengan}\{\begin{array}{c}\mu =8\\ \sigma =2\end{array}}\\ & ={\displaystyle \frac{1}{2\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left(\frac{x-8}{2}\right)}^2}}}}\\ & ={\displaystyle \frac{1}{\sqrt{8\pi }}{e}^{{.}^{-{\displaystyle \frac{(x-8{)}^2}{8}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika variabel acak}Z\text{berdistribusi normal}\\ & \text{N}(0,1),\text{nilai}\text{P}(Z<2)\text{adalah}....\\ & \text{a}.{\displaystyle {\int }_0^2{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\\ & \text{b}.{\displaystyle {\int }_2^{\mathrm{\infty }}{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\\ & \text{c}.{\displaystyle {\int }_{-\mathrm{\infty }}^2{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\\ & \text{d}.{\displaystyle {\int }_0^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{\text{x}-\mu }{\sigma }}\right)}^2}}}dz}}\\ & \text{e}.{\displaystyle {\int }_0^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{\text{x}-\mu }{\sigma }}\right)}^2}}}dz}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& P(\text{Z}<2),\text{Z}\sim \text{N}(0,1)\\ & =P(-\mathrm{\infty }<\text{Z}<0)+P(0<\text{Z}<2)\\ & ={\displaystyle {\int }_{-\mathrm{\infty }}^2{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika luas daerah di bawah kurva}\\ & \text{berdistribusi normal pada interval}\text{Z}>z\\ & \text{adalah}L,\text{nilai}\text{P}(-z<Z<z)\text{adalah}....\\ & \text{a}.{\displaystyle 0,5+L}\\ & \text{b}.0,5-L\\ & \text{c}.{\displaystyle 1-L}\\ & \text{d}.{\displaystyle 1-2L}\\ & \text{e}.{\displaystyle 2L}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}P& (-z<Z<z)\\ & =0,5-L+0,5-L\\ & =1-2L\\ & \text{Berikut ilustrasi kurva beserta luasnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}\text{X}\sim \text{N}(20,4)\text{dan}Z\sim N(0,1)\\ & \text{Jika}P(0<Z<1)=0,3413,\text{maka nilai}\\ & P(X<24)\text{adalah}....\\ & \text{a}.{\displaystyle 0,1587}\\ & \text{b}.{\displaystyle 0,3174}\\ & \text{c}.{\displaystyle 0,3413}\\ & \text{d}.{\displaystyle 0,6826}\\ & \text{e}.{\displaystyle 0,8413}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}X\sim N(20,4)\{\begin{array}{ll}\mu & =20\\ \sigma & =4\end{array}\\ & \text{Dan diketahui pula}P(0<Z<1)=0,3413\\ & \text{Jika}Z\sim N(0,1),\text{maka untuk}P(X<24)\\ & \text{transformasi}\text{x}=24\text{menjadi}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }=\frac{24-20}{4}=\frac{4}{4}=1}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}P(X<24)& =P(Z<1)\\ & =0,5+P(0<Z<1)\\ & =0,5+0,3413\\ & =0,8413\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai kuartil atas dari data}\\ & \text{berdistribusi normal baku}=q\\ & \text{Pernyataan berikut yang tepat adalah}....\\ & \text{a}.\text{Luas daerah pada}(Z<q)=0,25\\ & \text{b}.\text{Luas daerah pada}(Z>q)=0,25\\ & \text{c}.\text{Luas daerah pada}(0<Z<q)=0,25\\ & \text{d}.\text{Luas daerah pada}(Z<-0,25)=q\\ & \text{e}.\text{Luas daerah pada}(0<Z<0,25)=q\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \text{Pembahasan diserahkan kepada pembaca}\\ & \text{yang budiman}\end{array}$.

Koordinat Kartesius dan Koordinat Kutub serta Kuadran

C. Koordinat Kartesius dan Koordinat Kutub/Polar

Perhatikan ilustrasi berikut

$\begin{array}{|cc|}\hline \text{Kartesius}\to \text{Kutub}& \text{Kutub}\to \text{Kartesius}\\ P(x,y)\to P(r,{\alpha }^0)& P(r,{\alpha }^0)\to P(x,y)\\ \begin{array}{rl}& r=\sqrt{x^2+y^2},\\ & {\displaystyle \tan {\alpha }^0=\frac{y}{x},{\displaystyle {\alpha }^0=\arctan \frac{y}{x}}}\end{array}& \{\begin{array}{c}x=r.\cos {\alpha }^0\\ \\ y=r.\sin {\alpha }^0\end{array}\\ \hline\end{array}$.

D. Perbandingan Trigonometri di Berbagai Kuadran

D. 1 Untuk Sudut 0 sampai dengan 360 derajat.

$\begin{array}{cccccc}\text{Kuadran II}& & & & \text{Kuadran I}& \\ ({180}^{\circ }-\alpha )& & & & \text{Semua nilai trigon}& \mathbf{\text{positif}}\\ & & & & & \\ \text{Kuadran III}& & & & \text{Kuadran IV}& \\ ({180}^{\circ }+\alpha )& & & & ({360}^{\circ }-\alpha )& \end{array}$.

$\begin{array}{ll}(1).& \text{Perbandingan Trigonometri untuk Sudut}({90}^0-\alpha )\\ & \text{a}.\sin \alpha \to \sin ({90}^0-\theta )=\cos \theta \\ & \text{b}.\cos \alpha \to \cos ({90}^0-\theta )=\sin \theta \\ & \text{c}.\tan \alpha \to \tan ({90}^0-\theta )=\cot \theta \\ & \text{d}.\cot \alpha \to \cot ({90}^0-\theta )=\tan \theta \end{array}$.

$\begin{array}{ll}(2).& \text{Perbandingan Trigonometri untuk Sudut}({180}^0-\alpha )\\ & \text{a}.\sin \alpha \to \sin ({180}^0-\theta )=\sin \theta \\ & \text{b}.\cos \alpha \to \cos ({180}^0-\theta )=-\cos \theta \\ & \text{c}.\tan \alpha \to \tan ({180}^0-\theta )=-\tan \theta \\ & \text{d}.\cot \alpha \to \cot ({180}^0-\theta )=-\cot \theta \end{array}$.

$\begin{array}{ll}(3).& \text{Perbandingan Trigonometri untuk Sudut}({180}^0+\alpha )\\ & \text{a}.\sin \alpha \to \sin ({180}^0+\theta )=-\sin \theta \\ & \text{b}.\cos \alpha \to \cos ({180}^0+\theta )=-\cos \theta \\ & \text{c}.\tan \alpha \to \tan ({180}^0+\theta )=\tan \theta \\ & \text{d}.\cot \alpha \to \cot ({180}^0+\theta )=\cot \theta \end{array}$.

$\begin{array}{ll}(4).& \text{Perbandingan Trigonometri untuk Sudut}({360}^0-\alpha )\\ & \text{a}.\sin \alpha \to \sin ({360}^0-\theta )=-\sin \theta \\ & \text{b}.\cos \alpha \to \cos ({360}^0-\theta )=\cos \theta \\ & \text{c}.\tan \alpha \to \tan ({360}^0-\theta )=-\tan \theta \\ & \text{d}.\cot \alpha \to \cot ({360}^0-\theta )=-\cot \theta \end{array}$.

D. 2 Untuk Sudut yang Lain

$\begin{array}{rl}\text{a}.& \{\begin{array}{ll}\sin (-A)& =-\sin A\\ \cos (-A)& =\cos A\\ \tan (-A)& =-\tan A\end{array}\\ \text{b}.& \{\begin{array}{ll}\csc (-A)& =-\csc A\\ \sec (-A)& =\sec A\\ \cot (-A)& =-\cot A\end{array}\\ \text{c}.& \{\begin{array}{ll}\sin (n{.360}^{\circ }+A)& =\sin A\\ \cos (n{.360}^{\circ }+A)& =\cos A\\ \tan (n{.360}^{\circ }+A)& =\tan A\end{array},n\in \mathbb{N}\end{array}$.

Dengan catata: $0^{\circ }={360}^{\circ }={720}^{\circ }={1080}^{\circ }=n{.360}^{\circ }$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai}\\ & \text{a}.\sin {120}^{\circ }\\ & \text{b}.\cos {240}^{\circ }\\ & \text{c}.\tan {315}^{\circ }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.\sin {120}^{\circ }& =\sin ({180}^{\circ }-{60}^{\circ })=\sin {60}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{3},\text{atau}}\\ & =\sin ({90}^{\circ }+{30}^{\circ })=\cos {30}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.\cos {240}^{\circ }& =\cos ({180}^{\circ }+{60}^{\circ })=-\cos {60}^{\circ }\\ & =-{\displaystyle \frac{1}{2},\text{atau}}\\ & =\cos ({270}^{\circ }-{30}^{\circ })=-\sin {30}^{\circ }=-\frac{1}{2}\\ \text{c}.\tan {315}^{\circ }& =\tan ({360}^{\circ }-{45}^{\circ })=-\tan {45}^{\circ }\\ & =-1,\text{atau}\\ & =\tan ({270}^{\circ }+{45}^{\circ })=-\cot {45}^{\circ }=-1\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Buktikan bahwa}\\ \\ & \text{a}.{\displaystyle \frac{\cos ({90}^{\circ }-B)}{\sec B}+\frac{\sin ({90}^{\circ }-B)}{\csc B}=2\sin B\cos B}\\ \\ & \text{b}.\tan C+\tan ({90}^{\circ }-C)=\sec C.\sec ({90}^{\circ }-C)\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{\cos ({90}^{\circ }-B)}{\sec B}+\frac{\sin ({90}^{\circ }-B)}{\csc B}}\\ & ={\displaystyle \frac{\sin B}{\sec B}+\frac{\cos B}{\csc B}}\\ & ={\displaystyle \frac{\sin B}{{\displaystyle \frac{1}{\cos B}}}+\frac{\cos B}{{\displaystyle \frac{1}{\sin B}}}}\\ & =\sin B\cos B+\sin B\cos B\\ & =2\sin B\cos B◼\end{array}\\ & \begin{array}{rl}\text{b}.& \tan C+\tan ({90}^{\circ }-C)\\ & =\tan C+\cot C\\ & ={\displaystyle \frac{\sin C}{\cos C}+\frac{\cos C}{\sin C}}\\ & ={\displaystyle \frac{{\sin }^{2}C+{\cos }^{2}C}{\sin C\cos C}={\displaystyle \frac{1}{\sin C\cos C}}}\\ & ={\displaystyle \frac{1}{\cos C}.\frac{1}{\sin C}}\\ & =\sec C.\csc C\\ & =\sec C.\sec ({90}^{\circ }-C)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai}\\ & \text{a}.\tan (A-{90}^{\circ })\sin (-A)\\ & \text{b}.\cos {540}^{\circ }+\sin {690}^{\circ }\\ & \text{c}.\sin {2021}^{\circ }+\cos {2021}^{\circ }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \tan (A-{90}^{\circ })\sin (-A)\\ & =\tan (-({90}^{\circ }-A))(-\sin A)\\ & =-\tan ({90}^{\circ }-A)(-\sin A)\\ & =\tan ({90}^{\circ }-A)(\sin A)\\ & =\cot A.\sin A\\ & ={\displaystyle \frac{\cos A}{\sin A}.\sin A}\\ & =\cos A\end{array}\\ & \begin{array}{rl}\text{b}.& \cos {540}^{\circ }+\sin {690}^{\circ }\\ & =\cos ({360}^{\circ }+{180}^{\circ })+\sin ({720}^{\circ }-{30}^{\circ })\\ & =\cos (0^{\circ }+{180}^{\circ })+\sin (0^{\circ }-{30}^{\circ })\\ & =\cos {180}^{\circ }+\sin (-{30}^{\circ })\\ & =\cos {180}^{\circ }-\sin {30}^{\circ }\\ & =-1-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{3}{2}}\end{array}\\ & \begin{array}{rl}\text{c}.& \sin {2021}^{\circ }+\cos {2021}^{\circ }\\ & =\sin ({5.360}^{\circ }+{221}^{\circ })+\cos ({5.360}^{\circ }+{221}^{\circ })\\ & =\sin (0^{\circ }+{221}^{\circ })+\cos (0^{\circ }+{221}^{\circ })\\ & =\sin {221}^{\circ }+\cos {221}^{\circ }\\ & =\sin ({180}^{\circ }+{41}^{\circ })+\cos ({180}^{\circ }+{41}^{\circ })\\ & =-\sin {41}^{\circ }-\cos {41}^{\circ }\end{array}\end{array}$

Perbandingan Trigonometri Sudut dalam Segitiga Siku-Siku

B. Perbandingan Trigonometri Suatu Sudut

B. 1 Perbandingan Trigonometri pada Segitiha Siku-Siku

Perhatikanlah ilustrasi  dari segitiga ABC siku-siku di C berikut

$\begin{array}{c}\sin \alpha ={\displaystyle \frac{BC}{AB}\iff \csc \alpha ={\displaystyle \frac{AB}{BC}={\displaystyle \frac{1}{\sin \alpha }}}}\\ \\ \cos \alpha ={\displaystyle \frac{AC}{AB}\iff \sec \alpha ={\displaystyle \frac{AB}{AC}={\displaystyle \frac{1}{\cos \alpha }}}}\\ \\ \tan \alpha ={\displaystyle \frac{BC}{AC}\iff \cot \alpha ={\displaystyle \frac{AC}{BC}={\displaystyle \frac{1}{\tan \alpha }}}}\end{array}$.

B. 2 Perbandingan Trigonometri untuk Sudut Istimewa.

$\begin{array}{|ccccccccc|}\hline {\alpha }^0& 0^0& {30}^0& {45}^0& {60}^0& {90}^0& {180}^0& {270}^0& {360}^0\\ \sin {\alpha }^0& 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}& 1& 0& -1& 0\\ \cos {\alpha }^0& 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1& 0& 1\\ \tan {\alpha }^0& 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& TD& 0& TD& 0\\ \hline\end{array}$.

B. 3. Tripel Pythagoras pada Segitiga Siku-Siku

$\begin{array}{|cccc|}\hline a& b& c=\sqrt{a^2+b^2}& \text{Tripel}\\ 3& 4& 5& (3,4,5)\\ 5& 12& 13& (5,12,13)\\ 7& 24& 25& (7,24,25)\\ 8& 15& 17& (8,15,17)\\ 9& 40& 41& (9,40,41)\\ 11& 60& 61& (11,60,61)\\ 12& 35& 37& (12,35,37)\\ 15& 112& 113& (15,112,113)\\ 16& 63& 65& (16,63,65)\\ 17& 144& 145& (17,144,145)\\ 19& 180& 181& (19,180,181)\\ 20& 21& 29& (20,21,29)\\ 21& 220& 221& (21,220,221)\\ 29& 420& 421& (29,420,421)\\ 30& 224& 226& (30,224,226)\\ 31& 480& 481& (31,480,481)\\ 33& 544& 545& (33,544,545)\\ 35& 612& 613& (35,612,613)\\ 37& 684& 685& (37,684,685)\\ 39& 760& 761& (39,760,761)\\ 41& 840& 841& (41,840,841)\\ 43& 924& 925& (43,924,925)\\ 45& 1012& 1013& (45,1012,1013)\\ 47& 1104& 1105& (47,1104,1105)\\ 48& 55& 73& (48,55,73)\\ 49& 1200& 1201& (49,1200,1201)\\ 51& 1300& 1301& (51,1300,1301)\\ 60& 63& 87& (60,63,87)\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai perbandingan}\sin \alpha \\ & \cos \alpha ,\tan \alpha \text{untuk segitiga berikut}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Untuk sisi miringnya adalah}\\ & \begin{array}{rl}\text{Sisi miring}& =\sqrt{3^2+4^2}\\ & =\sqrt{9+16}\\ & =\sqrt{25}\\ & =5\end{array}\\ & \text{Sehingga}\\ & \sin \alpha ={\displaystyle \frac{3}{5},\cos \alpha =\frac{4}{5}\text{dan}\tan \alpha =\frac{3}{4}}\\ & \text{Dengan langkah sama}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Untuk sisi miringnya adalah}\\ & \begin{array}{rl}\text{Sisi miring}& =\sqrt{5^2+{12}^2}\\ & =\sqrt{25+144}\\ & =\sqrt{169}\\ & =13\end{array}\\ & \text{Sehingga}\\ & \sin \alpha =\frac{12}{13}\cos \alpha =\frac{5}{13}\text{dan}\tan \alpha =\frac{12}{5}.\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai dari}\\ & \begin{array}{ll}\text{a}.& (\tan {30}^0+\sin {30}^0)\cos {30}^0\\ \text{b}.& {(\tan {60}^0)}^2+4{(\sin {60}^0)}^2\\ \text{c}.& \tan {60}^0-\sin {60}^0-\tan {30}^0\\ \text{d}.& {\displaystyle \frac{1+\sin {30}^0}{\sin {30}^0}+\frac{\cos {30}^0}{1+\sin {30}^0}}\\ \text{e}.& {\displaystyle \frac{2\tan {30}^0}{1+{\tan }^2{30}^0}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\end{array}$.

$.\begin{array}{|ll|}\hline \begin{array}{ll}\text{a}.& (\tan {30}^0+\sin {30}^0)\cos {30}^0\\ & ={\displaystyle (\frac{1}{3}\sqrt{3}+\frac{1}{2}).\frac{1}{2}\sqrt{3}}\\ & ={\displaystyle \frac{1}{2}+\frac{1}{4}\sqrt{3}}\end{array}& \begin{array}{ll}\text{b}.& {(\tan {60}^0)}^2+4{(\sin {60}^0)}^2\\ & {\displaystyle ={\left(\sqrt{3}\right)}^2+4{\left(\frac{1}{2}\sqrt{3}\right)}^2}\\ & {\displaystyle =3+3}\\ & =6\end{array}\\ \begin{array}{ll}\text{c}.& \tan {60}^0-\sin {60}^0-\tan {30}^0\\ & ={\displaystyle \sqrt{3}-\frac{1}{2}\sqrt{3}-\frac{1}{3}\sqrt{3}}\\ & ={\displaystyle \sqrt{3}-\frac{5}{6}\sqrt{3}}\\ & ={\displaystyle \frac{1}{6}\sqrt{3}}\end{array}& \begin{array}{ll}\text{d}.& {\displaystyle \frac{1+\sin {30}^0}{\sin {30}^0}+\frac{\cos {30}^0}{1+\sin {30}^0}}\\ & ={\displaystyle \frac{1+\frac{1}{2}}{\frac{1}{2}}+\frac{\frac{1}{2}\sqrt{3}}{1+\frac{1}{2}}}\\ & ={\displaystyle 3+\frac{1}{3}\sqrt{3}}\end{array}.\\ \begin{array}{ll}\text{e}.& {\displaystyle \frac{2\tan {30}^0}{1+{\tan }^2{30}^0}}\\ & ={\displaystyle \frac{2\times \frac{1}{3}\sqrt{3}}{1+{\left(\frac{1}{3}\sqrt{3}\right)}^2}}\\ & ={\displaystyle \frac{\frac{2}{3}\sqrt{3}}{1+\frac{1}{3}}}\\ & ={\displaystyle \frac{\frac{2}{3}\sqrt{3}}{\frac{4}{3}}}\\ & ={\displaystyle \frac{1}{2}\sqrt{3}}\end{array}.& \\ \hline\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah ilustrasi berikut}\end{array}$

$.\begin{array}{rl}& \text{Jika Jarak antara kucing seorang pencatat}\\ & \text{dan kucing adalah 100 m, maka jarak}\\ & \text{Pencatat tersebut dengan seorang tentara}\\ & \text{sebagaimana gambar tersebut di atas adalah}?\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan gambar di atas dengan diberikan}\\ & \text{tambahan keterangan sebagai berikut}\end{array}$.

$.\begin{array}{rl}& \text{Ditanya berpakah panjang jarak }(x+100)?\\ & \begin{array}{rl}& \text{Pilih}:y=y\\ & \iff x.\tan {60}^0=(x+100).\tan {30}^0\\ & \iff x.\sqrt{3}=(x+100)\frac{1}{3}\sqrt{3}\\ & \iff 3x=x+100\\ & \iff 2x=100\\ & \iff x=50\end{array}\\ & \text{Jadi}x+100=50+100=150\text{meter}.\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah perbandingan trigonometri}\\ & \mathrm{\angle }XOA\text{jika}A(3,5)\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$\begin{array}{rl}& \text{Dengan memandang ilustrasi gambar di atas}\\ & \text{kita mendapatkan}\mathrm{△}OA{A}^{\prime },\text{dengan menggunakan}\\ & \text{teorema pythagoras kita mendapatkan}\\ & \begin{array}{rl}O{A}^2& ={\left(O{A}^{\prime }\right)}^2+{\left(A{A}^{\prime }\right)}^2\\ & =3^2+5^2\\ & =9+25\\ & =34\\ OA& =\sqrt{34}\end{array}.\\ & \text{Sehingga akan didapatkan}\\ & \begin{array}{ll}\text{a}.& {\displaystyle \sin A^{\prime }OA=\frac{5}{\sqrt{34}}=\frac{5}{34}\sqrt{34}}\\ \text{b}.& {\displaystyle \cos A^{\prime }OA=\frac{3}{\sqrt{34}}=\frac{3}{34}\sqrt{34}}\\ \text{c}.& {\displaystyle \tan A^{\prime }OA=\frac{5}{3}}\end{array}.\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai dari}\\ & \begin{array}{ll}\text{a}.& (\tan {60}^0+\sin {45}^0)\cos 0^0\\ \text{b}.& {(\tan {30}^0)}^2+4{(\sin {30}^0)}^2\\ \text{c}.& \tan {60}^0+\sin {60}^0+\tan {30}^0\\ \text{d}.& {\displaystyle \frac{1+\sin {45}^0}{\sin {30}^0}+\frac{\cos {30}^0}{1+\sin {45}^0}}\\ \text{e}.& {\displaystyle \frac{2\tan {60}^0}{1+{\tan }^2{60}^0}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan nilai perbandingan trigonometrinya}\\ & \text{dari gambar segitiga berikut}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui}\tan \alpha =0,75.\text{Tentukan nilai}\\ & \text{perbandingan trigonometri berikut}\\ & \begin{array}{ll}\text{a}.& \sin \alpha \\ \text{b}.& \cos \alpha \\ \text{c}.& \csc \alpha \\ \text{d}.& \sec \alpha \\ \text{e}.& \cot \alpha \\ \text{f}.& {\sin }^2\alpha +{\cos }^2\alpha \end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}1-\cos \beta ={\displaystyle \frac{9}{25(1+\cos \beta )}\text{Tentukan}}\\ & \text{nilai dari}\tan \beta \end{array}$.

DAFTAR PUSTAKA

1. Kurnianingsih, S., Kuntarti, Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 2 Standar Isi 2006. Jakarta: ESIS.
2. Marwanta, dkk. 2013. Matematika SMA Kelas X. Jakarta: YUDHISTIRA.
3. Yuana. R.A., Indriyastuti. 2017. Persepktif Matematika untuk Kelas X SMA dan MA Kelompok Matematika Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Trigonometri

 A. Ukuran Sudut

Sudut itu sendiri adalah suatu bangun yang dibatasi oleh dua buah sinar atau garis yang berpotongan di sekitar titik potongnya. Sebagai ilustrasinya perhatikanlah gambar berikut

Ada dua buah sinar yaitu  $\overrightarrow{OA}$  dan  $\overrightarrow{OB}$ yang bertemu atau berpotongan di titik $O$ den terbentuklah  $\mathrm{\angle }AOB$.

Sedangkan berkaitan dengan pengukuran sudut nantinya minimal kita mengenal 3 jenis, yaitu: derajat, radian, dan gone atau grade. Berikut hubungan ketiga jenis ikuran sudut yang dimaksud beserta lambang/notasi penulisannya. Untuk anak ditingkat MA/SMA ukuran sudut yang dikenalkan biasanya derajat dan radian, akan tetapi untuk anak SMK ada satu lagi yaitu satuan gone atau grade.

$\begin{array}{|c|}\hline 1\text{keliling}◯={360}^0=2\pi \text{radian}={400}^g\\ \text{atau}\\ {\displaystyle \frac{1}{2}\text{keliling}◯={180}^0=\pi \text{radian}={200}^9}\\ \text{boleh juga}\\ {\displaystyle \frac{1}{4}\text{keliling}◯={90}^0={\displaystyle \frac{1}{2}\pi \text{radian}={100}^9}}\\ \text{Jika ingin diperkecil lagi}\\ {\displaystyle \frac{1}{8}\text{keliling}◯={45}^0={\displaystyle \frac{1}{4}\pi \text{radian}={50}^9}}\\ \hline\end{array}$.

Perhatikan ilustrasi berikut

Sebagai catatan, terkadang sudut dalam satuan derajat dipresentasikan dalam bentuk seksagesimal, yaitu: diubah dalam satuan menit dan detik.

$\begin{array}{|l|}\hline \bullet {\displaystyle 1^{\circ }={60}^{\prime }={3600}^{″}}\\ \bullet 1^{\prime }={60}^{″}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Ubahlah sudut berikut ke satuan}\\ & \text{yang diminta}\\ & \text{a}.1^{\circ }=....\text{rad}\\ & \text{b}.1\text{rad}=....{}^0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \begin{array}{rl}\text{a}.{360}^0& =2\pi \text{rad}\\ 360\times 1^0& =2\pi \text{rad}\\ 1^0& ={\displaystyle \frac{2\pi }{360}}\\ & ={\displaystyle \frac{\pi }{180}\text{rad}}\end{array}\\ & \begin{array}{rl}\text{b}.2\pi \text{rad}& ={360}^0\\ 2\pi \times 1\text{rad}& ={360}^0\\ 1\text{rad}& ={\left({\displaystyle \frac{360}{2\pi }}\right)}^0\\ & ={\left({\displaystyle \frac{180}{\pi }}\right)}^0\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Ubahlah sudut berikut ke satuan }\\ & \text{yang diminta}\\ & \text{a}.53,{24}^{\circ }=....\text{(dalam sexagesimal)}\\ & \text{b}.{23}^{\circ }{12}^{\prime }{24}^{″}=....{}^0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.53,{24}^0& ={53}^0+0,{24}^0\\ & ={53}^0+0,24\times 1^0\\ & ={53}^0+0,24\times {60}^{\prime }\\ & ={53}^0+{14,4}^{\prime }\\ & ={53}^0+{14}^{\prime }+{0,4}^{\prime }\\ & ={53}^0+{14}^{\prime }+0,4\times 1^{\prime }\\ & ={53}^0+{14}^{\prime }+0,4\times {60}^{″}\\ & ={53}^0+{14}^{\prime }+{24}^{″}\\ 53,{24}^0& ={53}^0{14}^{\prime }{24}^{″}\end{array}\\ & \begin{array}{rl}\text{b}.{23}^0{12}^{\prime }{24}^{″}& ={23}^0+12\times 1^{\prime }+24\times 1^{″}\\ & ={23}^0+12\times {\left(\frac{1}{60}\right)}^0+24\times {\left(\frac{1}{3600}\right)}^0\\ & ={23}^0+0,2^0+0,00{\overline{666}}^0\\ & =23,20{\overline{666}}^0\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Ubahlah ke dalam sudut-sudut berikut dalam radian}\\ & \begin{array}{l}\\ \text{a}.& 3^0& \text{e}.& {90}^0& \text{i}.& {210}^0& \text{m}.& {300}^0\\ \text{b}.& {15}^0& \text{f}.& {120}^0& \text{j}.& {225}^0& \text{n}.& {315}^0\\ \text{c}.& {30}^0& \text{g}.& {135}^0& \text{k}.& {240}^0& \text{0}.& {12}^0{24}^{\prime }\\ \text{d}.& {60}^0& \text{h}.& {150}^0& \text{l}.& {270}^0& \text{p}.& {30}^0{12}^{\prime }{30}^{″}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat bahwa}:{180}^0=180\times 1^0=\pi rad\Rightarrow 1^0={\displaystyle \frac{\pi }{180}rad}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.3^0& =\cdots \pi rad\\ 3^0& ={\displaystyle \frac{3}{180}\pi rad}\\ & ={\displaystyle \frac{1}{60}\pi rad}\end{array}& \begin{array}{rl}\text{m}.{300}^0& =\cdots \pi rad\\ {300}^0& ={\displaystyle \frac{300}{180}\pi rad}\\ & ={\displaystyle \frac{5}{3}\pi rad}\end{array}\\ \begin{array}{rl}\text{c}.{30}^0& =\cdots \pi rad\\ {30}^0& ={\displaystyle \frac{30}{180}\pi rad}\\ & ={\displaystyle \frac{1}{6}\pi rad}\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{o}.{12}^0{24}^{\prime }& =\cdots \pi rad\\ {12}^0{24}^{\prime }& ={\displaystyle \frac{12+\left({\displaystyle \frac{24}{60}}\right)}{180}\pi rad}\\ & ={\displaystyle \frac{12+0,4}{180}\pi rad}\\ & ={\displaystyle \frac{12,4}{180}\pi rad}\\ & ={\displaystyle \frac{31}{450}\pi rad}\end{array}\\ \begin{array}{rl}\text{f}.{120}^0& =\cdots \pi rad\\ {120}^0& ={\displaystyle \frac{120}{180}\pi rad}\\ & ={\displaystyle \frac{2}{3}\pi rad}\end{array}& \\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Ubahlah ke dalam sudut-sudut berikut dalam derajat}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}\pi rad}& \text{e}.& {\displaystyle \frac{4}{3}\pi rad}\\ \\ \text{b}.& {\displaystyle \frac{3}{5}\pi rad}& \text{f}.& 2rad\\ \\ \text{c}.& {\displaystyle \frac{2}{9}\pi rad}& \text{g}.& 12rad\\ \\ \text{d}.& {\displaystyle \frac{7}{12}\pi rad}& \text{h}.& 15\pi rad\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat bahwa}:\pi rad={180}^0\Rightarrow 1rad={\displaystyle \frac{{180}^0}{\pi }}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.{\displaystyle \frac{1}{2}\pi rad}& ={\displaystyle \frac{1}{2}\pi \times {\displaystyle \frac{{180}^0}{\pi }}}\\ & ={90}^0\end{array}& \begin{array}{rl}\text{b}.{\displaystyle \frac{3}{5}\pi rad}& ={\displaystyle \frac{3}{5}\pi \times {\displaystyle \frac{{180}^0}{\pi }}}\\ & ={108}^0\end{array}\\ \begin{array}{rl}\text{d}.{\displaystyle \frac{7}{12}\pi rad}& ={\displaystyle \frac{7}{12}\pi \times {\displaystyle \frac{{180}^0}{\pi }}}\\ & ={105}^0\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{f}.2rad& =2\times {\displaystyle \frac{{180}^0}{\pi }}\\ & =2\times {\displaystyle \frac{{180}^0}{\frac{22}{7}}}\\ & ={\displaystyle \frac{{1260}^0}{11}}\\ & =114,\overline{54}{}^0\end{array}\\ \hline\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Ubahlah sudut berikut ke satuan }\\ & \text{yang diminta}\\ & \text{a}.{27}^0=.....\text{rad}\\ & \text{b}.28\text{rad}=...{.}^0\\ & \text{c}.31,{35}^{\circ }=....\text{(dalam sexagesimal)}\\ & \text{d}.{30}^{\circ }{24}^{\prime }{12}^{″}=....{}^0\end{array}$.

$\begin{array}{l}\\ 2.& \text{Ubahlah sudut berikut ke satuan }\\ & \text{yang diminta}\\ & \text{a}.{135}^0=.....\text{rad}\\ & \text{b}.6\text{rad}=...{.}^0\\ & \text{c}.23,{45}^{\circ }=....\text{(dalam sexagesimal)}\\ & \text{d}.{45}^{\circ }{50}^{\prime }{36}^{″}=....{}^0\end{array}$.

$\begin{array}{l}\\ 3.& \text{Silahkan dicoba sendiri soal yang belum}\\ & \text{dibahas}\end{array}$.

Vektor pada Ruang

 $\text{A. Vektor Di Ruang}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|cc|}\hline \text{Nama}& {\mathbf{\text{R}}}^3\\ \text{Vektor Satuan}& \text{Ruang (Bidang XYZ)}\\ {\hat{e}}_{\overline{a}}={\displaystyle \frac{\overline{a}}{\left|\overline{a}\right|}}& \{\begin{array}{ll}i=& \text{vektor satuan}\\ & \text{yang searah sumbu X}\\ j=& \text{Vektor satuan}\\ & \text{yang searah sumbu Y}\\ k=& \text{Vektor satuan}\\ & \text{searah sumbu Z}\end{array}\\ \text{Vektor nol}& \overrightarrow{O}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\\ \text{Vektor posisi}& \overrightarrow{OP}=\overrightarrow{p}=\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=p_1\overline{i}+p_2\overline{j}+p_3\overline{j}\\ \text{Besar Vektor}& \overrightarrow{OP}=\sqrt{p_1^2+p_2^2+p_3^2}\\ \hline\end{array}$

$\text{B. Operasi Vektor}$

$\text{1. Sifat-Sifat Aljabar Vektor}$

$\begin{array}{|lll|}\hline 1.& \text{Komutatif penjumlahan}& \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\\ 2.& \text{Asosiatif penjumlahan}& (\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{c}=\overrightarrow{a}+(\overrightarrow{b}+\overrightarrow{c})\\ 3.& \text{Elemen Identitas}& \overrightarrow{a}+\overrightarrow{0}=\overrightarrow{0}+\overrightarrow{a}=\overrightarrow{a}\\ 4.& \text{Invers Penjumlahan}& \overrightarrow{a}+(-\overrightarrow{a})=(-\overrightarrow{a})+\overrightarrow{a}=\overrightarrow{0}\\ 5.& \text{Perkalian dengan skalar}& k\left(l\overrightarrow{a}\right)=\left(kl\right)\overrightarrow{a}\\ & & k(\overrightarrow{a}+\overrightarrow{b})=k\overrightarrow{a}+k\overrightarrow{b}\\ & & k(\overrightarrow{a}-\overrightarrow{b})=k\overrightarrow{a}-k\overrightarrow{b}\\ 6.& \begin{array}{rl}& \text{Jika A, B, dan C segaris }\\ & \text{(Kolinear)}\end{array}& \{\begin{array}{l}\overrightarrow{AB}=k\overrightarrow{BC}\\ \overrightarrow{AC}=k\overrightarrow{AB}\\ dll\end{array}\\ \hline\end{array}$.

$\begin{array}{|cc|}\hline \text{Vektor}& \text{Contoh}\\ \overrightarrow{z}=a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}& \begin{array}{rl}& \text{diketahui}\overrightarrow{p}=\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}\\ & \text{maka pangjang vektor}\overrightarrow{p}\text{adalah}\\ & \left|\overrightarrow{p}\right|=\sqrt{1^2+(-2{)}^2+2^2}\\ & =\sqrt{1+4+4}=\sqrt{9}=3\end{array}\\ & \begin{array}{rl}& \text{Vektor satuan dari}\overrightarrow{p}\text{adalah}\\ & {\overrightarrow{e}}_{\overrightarrow{p}}=\frac{\overrightarrow{p}}{\left|\overrightarrow{p}\right|}={\displaystyle \frac{\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)}{3}}\\ & ={\displaystyle \frac{1}{3}\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)={\displaystyle \left(\begin{array}{c}\frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3}\end{array}\right)}}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui vektor-vektor}\overrightarrow{a}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\\ & \overrightarrow{b}=\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right),\text{dan}\overrightarrow{c}=\left(\begin{array}{c}7\\ 0\\ 4\end{array}\right),\\ & \text{tentukanlah hasil dari}\\ & \text{a}.\overrightarrow{a}+\overrightarrow{b}\\ & \text{b}.6\overrightarrow{a}+2\overrightarrow{b}\\ & \text{c}.2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ & \text{d}.{\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+{\displaystyle \frac{3}{4}\overrightarrow{b}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \overrightarrow{a}+\overrightarrow{b}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}2+(-3)\\ 1+(-5)\\ (-4)+2\end{array}\right)\\ & =\left(\begin{array}{c}2-3\\ 1-5\\ -4+2\end{array}\right)=\left(\begin{array}{c}-1\\ -4\\ -2\end{array}\right)\\ \text{b}.& 6\overrightarrow{a}+2\overrightarrow{b}=6\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+2\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}18-6\\ 6-10\\ -24+4\end{array}\right)=\left(\begin{array}{c}12\\ -4\\ -20\end{array}\right)\\ \text{c}.& 2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ & 2\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)-\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)+\left(\begin{array}{c}7\\ 0\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}4+3+7\\ 2+5+0\\ -8-2+4\end{array}\right)=\left(\begin{array}{c}17\\ 7\\ -6\end{array}\right)\\ \text{d}.& {\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+{\displaystyle \frac{3}{4}\overrightarrow{b}=\cdots }}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui vektor-vektor}\overrightarrow{a}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\\ & \text{tentukanlah}\left|\overrightarrow{a}\right|\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\left|\overrightarrow{a}\right|& =\sqrt{2^2+1^2+(-4{)}^2}\\ & =\sqrt{4+1+16}\\ & =\sqrt{21}\end{array}\end{array}$

$\text{2. Perkalian Skalar Dua Vektor}$

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

$\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$.

Misalkan diketahui

$\begin{array}{rl}& \overrightarrow{a}=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right),\overrightarrow{b}=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right),\text{maka}\\ & \overrightarrow{a}\cdot \overrightarrow{b}=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\cdot \left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)\\ & =a_1{b}_1+a_2{b}_2+a_3{b}_3\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Diketahui vektor-vektor}\overrightarrow{a}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\\ & \overrightarrow{b}=\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right),\text{dan}\overrightarrow{c}=\left(\begin{array}{c}7\\ 0\\ 4\end{array}\right),\\ & \text{tentukanlah hasil dari}\\ & \text{a}.\overrightarrow{a}\cdot \overrightarrow{b}\\ & \text{b}.\overrightarrow{a}\cdot \overrightarrow{c}\\ & \text{c}.\overrightarrow{b}\cdot \overrightarrow{c}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \overrightarrow{a}\cdot \overrightarrow{b}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\cdot \left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\\ & =(2)(-3)+(1)(-5)+(-4)(2)\\ & =-6-5-8=-19\\ \text{b}& \overrightarrow{a}\cdot \overrightarrow{c}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\cdot \left(\begin{array}{c}7\\ 0\\ 4\end{array}\right)\\ & =(2)(7)+(1)(0)+(-4)(4)\\ & =14+0-16=-2\\ \text{c}& \overrightarrow{b}\cdot \overrightarrow{c}=\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\cdot \left(\begin{array}{c}7\\ 0\\ 4\end{array}\right)\\ & =(-3)(7)+(-5)(0)+(2)(4)\\ & =-21+0+8=-13\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai}t\text{jika}\\ & \overrightarrow{p}=3\overline{i}+t\overline{j}+\overline{k}\text{dan}\overrightarrow{p}\cdot \overrightarrow{p}=13\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \overrightarrow{p}\cdot \overrightarrow{p}=13\\ & \overrightarrow{p}\cdot \overrightarrow{p}=\left|\overrightarrow{p}\right|\left|\overrightarrow{p}\right|\cos 0^{\circ }=13,\\ & \text{ingat bahwa}\mathrm{\angle }(\overrightarrow{p},\overrightarrow{p})=0^{\circ }\\ & \text{dan nilai}\cos 0^{\circ }=1,\\ & \text{maka}\\ & \overrightarrow{p}\cdot \overrightarrow{p}={\left|\overrightarrow{p}\right|}^2.1=13\iff {\left|\overrightarrow{p}\right|}^2=13\\ & \iff {\left(\sqrt{3^2+t^2+1^2}\right)}^2=13\\ & \iff 3^2+t^2+1^2=13\\ & \iff 9+t^2+1=13\\ & \iff t^2=13-9-1=10\\ & \iff t^2=3\\ & \iff t=\pm \sqrt{3}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Diketahui}\overrightarrow{p}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}4\\ -1\\ t\end{array}\right)\\ & \text{Jika}\overrightarrow{p}\text{tegak lurus}\overrightarrow{q}\text{maka}\\ & \text{tentukanlah nilai}t\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \overrightarrow{p}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}4\\ -1\\ t\end{array}\right)\\ & \text{dengan}\overrightarrow{p}\text{dan}\overrightarrow{q}\text{tegak lurus}\\ & \text{artinya}\mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})={90}^{\circ }.\text{Sehingga}\\ & \text{nilai}\cos {90}^{\circ }=0\\ & \text{maka}\\ & \overrightarrow{p}\cdot \overrightarrow{q}=\left|\overrightarrow{p}\right|\left|\overrightarrow{q}\right|\cos \theta \\ & \overrightarrow{p}\cdot \overrightarrow{q}=\left|\overrightarrow{p}\right|\left|\overrightarrow{q}\right|\cdot 0=0\\ & \iff \left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\cdot \left(\begin{array}{c}4\\ -1\\ t\end{array}\right)=0\\ & \iff (-2)(4)+(1)(-1)+(3)(t)=0\\ & \iff -8-1+3t=0\\ & \iff 3t=9\\ & \iff t=3\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}\left|\bar{u}\right|=6,\left|\bar{v}\right|=4\sqrt{3},\text{dan}|\bar{u}-\bar{v}|=8\\ & \text{tentukanlah nilai dari}\\ & \text{a}.\bar{u}\bullet \bar{v}\\ & \text{b}.|\bar{u}+\bar{v}|\\ & \text{c}.\mathbf{\text{cosinus}}\text{sudut antara}\bar{u}\text{dan}\bar{v}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \bar{u}\bullet \bar{v}=\cdots \\ & 2.\bar{u}\bullet \bar{v}={\left|\bar{u}\right|}^2+{\left|\bar{v}\right|}^2-{|\bar{u}-\bar{v}|}^2\\ & 2.\bar{u}\bullet \bar{v}=6^2+(4\sqrt{3}{)}^2-8^2\\ & 2.\bar{u}\bullet \bar{v}=36+48-64=84-64=20\\ & \bar{u}\bullet \bar{v}={\displaystyle \frac{20}{2}=10}\\ \text{b}.& {|\bar{u}+\bar{v}|}^2={\left|\bar{u}\right|}^2+{\left|\bar{v}\right|}^2+2.\bar{u}\bullet \bar{v}\\ & {|\bar{u}+\bar{v}|}^2=6^2+(4\sqrt{3}{)}^2+20\\ & =84+20=104\\ & |\bar{u}+\bar{v}|=\sqrt{104}\\ \text{c}.& \cos \mathrm{\angle }(\bar{u},\bar{v})={\displaystyle \frac{\bar{u}\bullet \bar{v}}{\left|\bar{u}\right|.\left|\bar{v}\right|}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}}\\ & ={\displaystyle \frac{10\sqrt{3}}{72}=\frac{5}{36}\sqrt{3}}\\ & \mathbf{\text{Berikut ilustrasi gambarnya}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Diketahui}\left|\overrightarrow{a}\right|=\sqrt{3},\left|\overrightarrow{b}\right|=1,\text{dan}|\overrightarrow{a}-\overrightarrow{b}|=1\\ & \text{maka panjang vektor}\overrightarrow{a}+\overrightarrow{b}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{3}& & & \text{d}.& 2\sqrt{2}\\ \text{b}.& \sqrt{5}& \text{c}.& \sqrt{7}& \text{e}.& 3\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Diketahui}& \text{sebagaimana pada soal}\\ {|\overrightarrow{a}-\overrightarrow{b}|}^2& ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ 1^2& ={\left(\sqrt{3}\right)}^2+1^2-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta & =3\\ \text{maka pan}& \text{jang vektor}\overrightarrow{a}+\overrightarrow{b}\text{adalah}\\ |\overrightarrow{a}+\overrightarrow{b}|& =\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }\\ & =\sqrt{{\left(\sqrt{3}\right)}^2+1^2+3}\\ & =\sqrt{3+1+3}\\ & =\sqrt{7}\end{array}\end{array}$

RANGKUMAN 

Klik di sini

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA. 
3. Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Contoh Soal 8 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 36.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_0^{n^2}\left[\sqrt{x}\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{(n+1)n(n+1)}{6}}\\ \text{b}.{\displaystyle \frac{n(n-1)(n-2)}{6}}\\ \text{c}.{\displaystyle \frac{n(n+1)(2n+1)}{6}}\\ \text{d}.{\displaystyle \frac{n(n-1)(3n+1)}{6}}\\ \text{e}.{\displaystyle \frac{n(n-1)(4n+1)}{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_0^{n^2}\left[\sqrt{x}\right]dx}\\ & ={\displaystyle {\int }_0^1\left[\sqrt{0}\right]dx+{\int }_1^{2^2}\left[\sqrt{1}\right]dx+{\int }_{2^2}^{3^2}\left[\sqrt{2^2}\right]dx}\\ & +{\displaystyle {\int }_{3^2}^{4^2}\left[\sqrt{3^2}\right]dx+\cdots +{\int }_{(n-1{)}^2}^{n^2}\left[\sqrt{(n-1{)}^2}\right]dx}\\ & ={\displaystyle {\int }_0^{1}0dx+{\int }_1^{4}1dx+{\int }_4^{9}2dx+{\int }_9^{16}3dx}\\ & +\cdots +{\int }_{(n-1{)}^2}^{n^2}(n-1)dx\\ & =0+x{|}_1^4+2x{|}_4^9+3x{|}_9^{16}+\cdots +(n-1)x{|}_{(n-1{)}^2}^{n^2}\\ & =0+(4-1)+2(9-4)+3(16-9)+\cdots +(n-1)(2n-1)\\ & =3+2.5+3.7+4.9+5.11+\cdots +(n-1)(2n-1)\\ & =3+10+21+36+55+\cdots +(n-1)(2n-1)\\ & (\text{Deretnya berupa barisan aritmetika tingkat 2})\\ & ={\displaystyle \frac{n(n-1)(4n+1)}{6}}\end{array}$.

DAFTAR PUSTAKA

1. Hutahaean. 1980. Kalkulus Diferensial dan Integral I. Jakarta: GRAMEDIA.
2. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007.  Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar ISI 2006. Jakarta: ESIS.
3. Nugroho, P.A., Gunarto, D. 2013. Big Bank Soal + Bahas Matematika SMA/MA Kelas 1,2,3. Jakarta: WAHYUMEDIA.
4. Sukino. 2015. Matematika Kelompok Peminatan Matematika dan Ilmu Alam. Jakarta: ERLANGGA.
5. Suparmin, S., Malau, A. 2014. Seri Kinomatika: Mainstream Matematika Dasar & Matematika IPA untuk Siswa SMA/MA Kelompok MIA. Bandung: YRAMA WIDYA.

Youtube:

Calculus | Find the Integral of the Floor Function of x

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Contoh Soal 7 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 31.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{2}}]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 3}\\ \text{b}.{\displaystyle 3\frac{1}{2}}\\ \text{c}.{\displaystyle 4}\\ \text{d}.{\displaystyle 4\frac{1}{2}}\\ \text{e}.{\displaystyle 5}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{2}}]dx}\\ & ={\displaystyle {\int }_{-1}^{-\frac{1}{2}}-1dx+{\int }_{-\frac{1}{2}}^{\frac{1}{2}}0dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}1dx+{\int }_{\frac{3}{2}}^{\frac{5}{2}}2dx+{\int }_{\frac{5}{2}}^{3}3dx}\\ & =-1{|}_{-1}^{-\frac{1}{2}}+0{|}_{-\frac{1}{2}}^{\frac{1}{2}}+x{|}_{\frac{1}{2}}^{\frac{3}{2}}+2x{|}_{\frac{3}{2}}^{\frac{5}{2}}+3x{|}_{\frac{5}{2}}^3\\ & =-(-{\displaystyle \frac{1}{2}+1})+0+\left({\displaystyle \frac{3}{2}-\frac{1}{2}}\right)+2(\frac{5}{2}-\frac{3}{2})+3(3-{\displaystyle \frac{5}{2}})\\ & =-{\displaystyle \frac{1}{2}+0+1+2+\frac{3}{2}}\\ & ={\displaystyle 4}\\ & \text{Perhatikan tabel berikut}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{3}}]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 3}\\ \text{b}.{\displaystyle 3\frac{1}{3}}\\ \text{c}.{\displaystyle 4\frac{2}{3}}\\ \text{d}.{\displaystyle 5\frac{1}{3}}\\ \text{e}.{\displaystyle 6\frac{1}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{3}}]dx}\\ & ={\displaystyle {\int }_{-1}^{-\frac{1}{3}}-1dx+{\int }_{-\frac{1}{3}}^{\frac{2}{3}}0dx+{\int }_{\frac{2}{3}}^{\frac{5}{3}}1dx+{\int }_{\frac{5}{3}}^{\frac{8}{3}}2dx+{\int }_{\frac{8}{3}}^{3}3dx}\\ & =-1{|}_{-1}^{-\frac{1}{3}}+0{|}_{-\frac{1}{3}}^{\frac{2}{3}}+x{|}_{\frac{2}{3}}^{\frac{5}{3}}+2x{|}_{\frac{5}{3}}^{\frac{8}{3}}+3x{|}_{\frac{8}{3}}^3\\ & =-(-{\displaystyle \frac{1}{3}+1})+0+\left({\displaystyle \frac{5}{3}-\frac{2}{3}}\right)+2(\frac{8}{3}-\frac{5}{3})+3(3-{\displaystyle \frac{8}{3}})\\ & =-{\displaystyle \frac{2}{3}+0+1+2+\frac{3}{3}}\\ & ={\displaystyle 3\frac{1}{3}}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_0^n\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle n^2}\\ \text{b}.{\displaystyle \frac{n(n-1)}{2}}\\ \text{c}.{\displaystyle \frac{n(n+1)}{2}}\\ \text{d}.{\displaystyle \frac{(n+1)(n+2)}{2}}\\ \text{e}.{\displaystyle \frac{n^2+1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_0^{n}f(x)dx={\displaystyle {\int }_0^n\left[x\right]dx}}\\ & ={\displaystyle {\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx+\cdots +{\int }_{n-1}^n(n-1)dx}\\ & =0+x{|}_1^2+2x{|}_2^3+\cdots +(n-1)x{|}_{n-1}^n\\ & =0+(2-1)+2(3-2)+\cdots +(n-1)(n-(n-1))\\ & =0+1+2+\cdots (n-1)\\ & ={\displaystyle \frac{(n-1)((n-1)+1)}{2}}\\ & ={\displaystyle \frac{n(n-1)}{2}}\\ & \text{Perhatikan tabel berikut}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-2022}^{2022}\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left[x\right]dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left[x\right]dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_{-2022}^{2022}\left[x\right]dx}\\ & ={\displaystyle {\int }_{-2022}^{-2021}\left[x\right]dx+{\displaystyle {\int }_{-2021}^{-2020}\left[x\right]dx+\cdots +{\int }_{2021}^{2022}\left[x\right]dx}}\\ & =-2022+(-2021)+\cdots +2020+2021\\ & =-2022\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-2022}^{2022}(x-\left[x\right])dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left[x\right]dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left[x\right]dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_{-2022}^{2022}(x-\left[x\right])dx}\\ & ={\displaystyle {\int }_{-2022}^{2022}xdx-\left({\displaystyle {\int }_{-2022}^{-2021}\left[x\right]dx+\cdots +{\int }_{2021}^{2022}\left[x\right]dx}\right)}\\ & \text{Sebelumnya perlu diingat bahwa}\\ & {\displaystyle {\int }_{-2022}^{2022}xdx\text{adalah}\text{fungsi ganjil, maka}=0}\\ & \text{Sehingga nilai akhirnya adalah}\\ & =0-(-2022)\text{lihat pembahasan sebelumnya}\\ & =2022\end{array}\end{array}$.