Lanjutan 2 Contoh Soal Persiapan untuk Numerasi AN BK 2022

$\text{CONTOH SOAL 21}$.

$\begin{array}{l}\\ & \text{Nilai eksak dari}\sin {36}^{\circ }\text{adalah}....\\ & \begin{array}{|c|}\hline \bullet \sin 2\theta =2\sin \theta \cos \theta \\ \bullet \cos 2\theta =2{\cos }^2\theta -1\\ \bullet {\sin }^2\theta +{\cos }^2\theta =1\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}}}& & & \text{d}.& {\displaystyle \frac{\sqrt{5}-1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}& & & \text{e}.& {\displaystyle \frac{\sqrt{5}-1}{2}}\\ \text{c}.& {\displaystyle {\displaystyle \frac{\sqrt{5}+1}{4}}}\end{array}\end{array}$.

$\text{SOLUSI SOAL 21}$.

Perhatikanlah ilustrasi segitiga berikut ini

$.\begin{array}{rl}& \text{Perhatikan bahwa}△ABC\text{sama kaki}\\ & \text{dengan}AD=DC=CB=1,AC=x\\ & \text{Diketahui pula}CD\text{adalah garis bagi}\\ & \text{serta}ABC\text{sebangun}△BCD\\ & \text{akibatnya}:\\ & \text{perbandingan sisi yang bersesuaian}\\ & \text{akan sama},\text{maka}\\ & {\displaystyle \frac{AB}{BC}={\displaystyle \frac{BC}{AB-AD}}}\\ & \iff {\displaystyle \frac{x}{1}=\frac{1}{x-1}}\\ & \iff x(x-1)=1\\ & \iff x^2-x-1=0\\ & \iff x={\displaystyle \frac{1\pm \sqrt{5}}{2}}\\ & \text{akibatnya}AB=AC={\displaystyle \frac{1+\sqrt{5}}{2}}\\ & \text{Selanjutnya gunakan}\text{aturan sinus}\\ & {\displaystyle \frac{AB}{\sin \mathrm{\angle }C}=\frac{BC}{\sin \mathrm{\angle }A}}\\ & \iff {\displaystyle \frac{AB}{BC}=\frac{\sin \mathrm{\angle }C}{\sin \mathrm{\angle }A}}\\ & \iff {\displaystyle \frac{\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)}{1}=\frac{\sin {72}^{\circ }}{\sin {36}^{\circ }}}\\ & \iff {\displaystyle \frac{1+\sqrt{5}}{2}={\displaystyle \frac{2\sin {36}^{\circ }\cos {36}^{\circ }}{\sin {36}^{\circ }}}}\\ & \iff {\displaystyle \frac{1+\sqrt{5}}{2}=2\cos {36}^{\circ }}\\ & \iff \cos {36}^{\circ }=\iff {\displaystyle \frac{1+\sqrt{5}}{4}}\\ & \text{Dari fakta di atas kita akan dengan}\\ & \text{mudah menentukan nilai sinusnya}\\ & \text{yaitu dengan menggunakan}\\ & \text{identitas trigonometri berikut}:\\ & {\sin }^2{36}^{\circ }+{\cos }^2{36}^{\circ }=1\\ & \iff {\sin }^2{36}^{\circ }=1-{\cos }^2{36}^{\circ }\\ & \iff \sin {36}^{\circ }=\sqrt{1-{\cos }^2{36}^{\circ }}\\ & \iff =\sqrt{1-{\left({\displaystyle \frac{1+\sqrt{5}}{4}}\right)}^2}\\ & \iff =\sqrt{1-{\displaystyle \frac{6+2\sqrt{5}}{16}}}\\ & \iff =\sqrt{{\displaystyle \frac{10-2\sqrt{5}}{16}}}\\ & \iff ={\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}\end{array}$.

Untuk menjawab soal no.22,23,24, 25 dan 26 perlu diketahui bahwa

$\lfloor x\rfloor =\text{bilangan bulat terbesar}\le x$.

$\lfloor 3,14\rfloor =3,\lfloor -7,2\rfloor =-8,\lfloor 2\rfloor =2$.

$\text{CONTOH SOAL 22}$.

$\begin{array}{rl}& \text{Jika}\lfloor a\rfloor \times a=68\text{dan}\lfloor b\rfloor \times b=109,\\ & \text{maka nilai dari}\lfloor a\rfloor \times \lfloor b\rfloor -\lfloor a+b\rfloor =\cdots \end{array}$.

$\text{SOLUSI SOAL 22}$.

$\begin{array}{rl}& \text{Perhatikan bahwa}\\ & \lfloor a\rfloor a=68\text{dan}\lfloor b\rfloor b=109\\ & \iff {\lfloor a\rfloor }^2\le \lfloor a\rfloor a\text{dan}{\lfloor b\rfloor }^2\le \lfloor b\rfloor b\\ & \iff {\lfloor a\rfloor }^2\le 68\text{dan}{\lfloor b\rfloor }^2\le 109\\ & \iff \lfloor a\rfloor =8\text{dan}\lfloor a\rfloor =10\\ & \text{Selanjutnya}\\ & \bullet \lfloor a\rfloor a=68\Rightarrow 8a=68\Rightarrow a=8,5\\ & \bullet \lfloor b\rfloor b=109\Rightarrow 10b=109\Rightarrow b=10,9\\ & \text{Sehingga nilai}\\ & \lfloor a\rfloor \times \lfloor b\rfloor -\lfloor a+b\rfloor \\ & =8\times 10-\lfloor 8,5+10,9\rfloor \\ & =80-\lfloor 19,4\rfloor \\ & =80-19\\ & =61\end{array}$.

$\text{CONTOH SOAL 23}$.

$\begin{array}{rl}& \text{Jika}a\lfloor a\rfloor =17\text{dan}b\lfloor b\rfloor =11,\text{maka}\\ & \text{nilai dari}a-b=\cdots \end{array}$.

$\text{SOLUSI SOAL 23}$.

$\begin{array}{rl}& \text{Perhatikan bahwa}\\ & a\lfloor a\rfloor =17\text{dan}b\lfloor b\rfloor =11\\ & \iff {\lfloor a\rfloor }^2\le a\lfloor a\rfloor \text{dan}{\lfloor b\rfloor }^2\le b\lfloor b\rfloor \\ & \iff {\lfloor a\rfloor }^2\le 17\text{dan}{\lfloor b\rfloor }^2\le 11\\ & \iff \lfloor a\rfloor =4\text{dan}\lfloor a\rfloor =3\\ & \text{Selanjutnya}\\ & \bullet a\lfloor a\rfloor =17\Rightarrow 4a=17\Rightarrow a={\displaystyle \frac{17}{4}}\\ & \bullet b\lfloor b\rfloor =11\Rightarrow 3b=11\Rightarrow b={\displaystyle \frac{11}{3}}\\ & \text{Sehingga nilai}\\ & a-b={\displaystyle \frac{17}{4}-\frac{11}{3}}\\ & ={\displaystyle \frac{51-44}{12}}\\ & ={\displaystyle \frac{7}{12}}\end{array}$.

$\text{CONTOH SOAL 24}$.

$\begin{array}{rl}& \text{Jika}\lfloor x\rfloor +\lfloor y\rfloor +y=43,8\text{dan}\\ & x+y-\lfloor x\rfloor =18,4,\text{maka nilai}\\ & \text{dari}10(x+y)=\cdots \end{array}$.

$\text{SOLUSI SOAL 24}$.

$\begin{array}{rl}& \text{Diketahui bahwa}\\ & \bullet \lfloor x\rfloor +\lfloor y\rfloor +y=43,8\text{dan}\\ & \bullet x+y-\lfloor x\rfloor =18,4\\ & \text{Sekarang misalkan untuk}\\ & \bullet a\le x<a+1\Rightarrow \lfloor x\rfloor =a\\ & \bullet b\le y\le b+1\Rightarrow \lfloor y\rfloor =b\\ & \text{Selanjutnya untuk}x\text{dan}y\\ & \text{dapat kita nyatakan dengan}\\ & \bullet x=a+m,\text{dengan}0\le m<1\\ & \bullet y=b+n,\text{dengan}0\le n<1\\ & \begin{array}{rl}& \mathbf{\text{Untuk persamaan pertama}}\\ & \lfloor x\rfloor +\lfloor y\rfloor +y=43,8\\ & \iff a+b+b+n=43,8\\ & \iff a+2b+n=43,8\\ & \text{didapatkan}a+2b=43,\text{dan}n=0,8\\ & \mathbf{\text{Untuk persamaan kedua}}\\ & x+y-\lfloor x\rfloor =18,4\\ & \iff a+m+b+n-a=18,4\\ & \iff b+m+n=18,4\\ & \iff b+m+0,8=18,4\\ & \iff b+m=17,6\\ & \text{didapatkan}b=17,\text{dan}m=0,6\\ & \mathbf{\text{Selanjutnya perhatikan bahwa}}\\ & a+2b=43\iff a+2.(17)=43\\ & \iff a+34=43\iff a=43-34=9\\ & \mathbf{\text{Sehingga kita akan mendapatkan nilai}}\\ & \bullet x=a+m=9+0,6=9,6\\ & \bullet y=b+n=17+0,8=17,8\\ & \text{Jadi},10(x+y)=10(9,6+17,8)=274\end{array}\end{array}$.

$\text{CONTOH SOAL 25}$.

$\begin{array}{rl}& \text{Jika}\lfloor 3x+y\rfloor =12\text{dan}\lfloor x+3y\rfloor =14,\\ & \text{maka nilai}\lfloor x+y\rfloor =\cdots \end{array}$.

$\text{SOLUSI SOAL 25}$.

$\begin{array}{rl}& \text{Diketahui bahwa}\\ & \bullet \lfloor 3x+y\rfloor =12\iff 12\le 3x+y<13\cdots (1)\\ & \bullet \lfloor x+3y\rfloor =14\iff 14\le x+3y<15\cdots (2)\\ & \text{Jika kedua ketaksamaan dijumlahkan, maka}\\ & 26\le 4x+4y<28\iff {\displaystyle \frac{26}{4}\le x+y<{\displaystyle \frac{28}{4}}}\\ & \iff 6{\displaystyle \frac{1}{2}\le x+y<7}\\ & \text{Jadi},\lfloor x+y\rfloor =6\end{array}$.

$\text{CONTOH SOAL 26}$.

$\begin{array}{rl}& \text{Jika}\lfloor {\displaystyle \frac{3x-7}{5}}\rfloor ={\displaystyle \frac{x}{3},\text{maka untuk}x\text{bulat terkecil}=\cdots }\end{array}$.

$\text{SOLUSI SOAL 26}$.

$\begin{array}{rl}& \text{Diketahui bahwa}\\ & \lfloor {\displaystyle \frac{3x-7}{5}}\rfloor ={\displaystyle \frac{x}{3}}\\ & \iff {\displaystyle \frac{x}{3}\le {\displaystyle \frac{3x-7}{5}<{\displaystyle \frac{x}{3}+1}}}\\ & \iff {\displaystyle \frac{x}{3}-\frac{x}{3}\le {\displaystyle \frac{3x-7}{5}-\frac{x}{3}<{\displaystyle \frac{x}{3}-\frac{x}{3}+1}}}\\ & \iff 0\le {\displaystyle \frac{4x-21}{15}<1}\\ & \iff 0\le 4x-21<15\\ & \iff 0+21\le 4x-21+21<15+21\\ & \iff 21\le 4x<36\\ & \iff {\displaystyle \frac{21}{4}\le x<9}\\ & \text{Jadi},x=6\end{array}$.

DAFTAR PUSTAKA

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