Aturan Sinus dan Aturan Cosinus

A. Pendahuluan

Aturan sinus maupun aturan cosinus keduanya sangat bermanfaat berkaitan dengan unsur segitiga baik siku-siku maupun segitiga bebas dalam penentuan besar sudut dalam segitiga tersebut maupun panjang sisi yang diingin. Dalam hal penentuan besar sudut atau menentukan permasalahan panjang salah satu sisi segitiga jika nantinya sudut diketahui, terkadang besar sudutnya tidak cuma lancip, dibanyak soal dimunculkan sudut tumpul. Oleh karenanya ada baiknya pembaca mengetahui nilai perbandingan trigonometri diberbagai kuadran dan nilai sudut-sudut istimewa dalam trigonometri serta tak lupa juga beberapa identitas trigonometri.

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

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

$\begin{array}{rl}& \text{Macam-Macam Identitas Trigonometri Dasar}\\ & 1.\csc \alpha ={\displaystyle \frac{1}{\sin \alpha }5.\tan \alpha ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}}\\ & 2.\sec \alpha ={\displaystyle \frac{1}{\cos \alpha }6.{\tan }^2\alpha +1={\sec }^2\alpha }\\ & 3.\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }7.{\cot }^2\alpha +1={\csc }^2\alpha }\\ & 4.\cot \alpha ={\displaystyle \frac{\cos \alpha }{\sin \alpha }8.{\sin }^2\alpha +{\cos }^2=1}\end{array}$.

B. Aturan Sinus

$\begin{array}{|c|}\hline {\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R}\\ \hline\end{array}$.

C. Aturan Cosinus

$\begin{array}{|c|}\hline \begin{array}{rl}\bullet & \cos \mathrm{\angle }A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ \bullet & \cos \mathrm{\angle }B={\displaystyle \frac{a^2+c^2-a^2}{2ac}}\\ \bullet & \cos \mathrm{\angle }C={\displaystyle \frac{a^2+b^2-a^2}{2ab}}\end{array}\\ \hline\end{array}$.

D. Luas Segitiga

$\begin{array}{rl}\mathbf{\text{Luas}}\mathrm{△}ABC& =\frac{1}{2}bc.\sin \mathrm{\angle }A\\ & =\frac{1}{2}ac.\sin \mathrm{\angle }B\\ & =\frac{1}{2}ab.\sin \mathrm{\angle }C\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\mathrm{△}ABC\text{dengan panjang sisi}\\ & AC=10cm\text{dan}BC=16cm\text{serta luas}\\ & \mathrm{△}ABC=40c{m}^2,\text{maka besar}\mathrm{\angle }ACB\\ & \text{jika sudutnya lancip adalah}\cdots \\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\{\begin{array}{c}AC=10cm\\ BC=16cm\\ L_{\mathrm{△}}=40c{m}^2\end{array},\text{maka}\\ & \begin{array}{rl}{L}_{\mathrm{△}ABC}& =\frac{1}{2}.AC.BC.\sin \mathrm{\angle }ACB\\ 40& =\frac{1}{2}.10.16.\sin \mathrm{\angle }ACB\\ 40& =80.\sin \mathrm{\angle }ACB\\ \frac{40}{80}& =\sin \mathrm{\angle }ACB\\ \sin \mathrm{\angle }ACB& =\frac{1}{2}\\ \sin \mathrm{\angle }ACB& =\sin {30}^0\\ \mathrm{\angle }ACB& ={30}^0\end{array}.\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Jika }\\ & AB+3=BC+2=CD+1=AD=4cm,\\ & \text{maka}\cos \mathrm{\angle }BAD\text{adalah}\cdots \\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan kembali ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Langkah awal kita gunakan garis bantu BD}\\ & \text{untuk nantinya kita mendapatkan nilai}cos\\ & \text{dari sudut A, yaitu}:\\ & \begin{array}{rl}B{D}^2& =B{A}^2+D{A}^2-2.BA.DA.\cos \mathrm{\angle }A\\ & =1^2+4^2-2.1.4.\cos \mathrm{\angle }A\\ & =17-8\cos \mathrm{\angle }A\\ B{D}^2& =B{C}^2+D{C}^2-2.BC.DC.\cos \mathrm{\angle }C\\ & =2^2+3^2-2.2.3.\cos \mathrm{\angle }C\\ & =13-12\cos \mathrm{\angle }C\end{array}\\ & \text{Perlu diketahui bahwa}\\ & \mathrm{\angle }A+\mathrm{\angle }C=\mathrm{\angle }B+\mathrm{\angle }C={180}^0\\ & \text{karena ABCD segiempat talibusur, sehingga}\\ & \mathrm{\angle }C={180}^0-\mathrm{\angle }A\\ & \begin{array}{rl}B{D}^2& =B{D}^2\\ 17-8\cos \mathrm{\angle }A& =13-12\cos \mathrm{\angle }C\\ 12\cos \mathrm{\angle }C-8\cos \mathrm{\angle }A& =13-17\\ 12(\cos ({180}^0-\mathrm{\angle }A))-8\cos \mathrm{\angle }A& =-4\\ 12(-\cos \mathrm{\angle }A)-8\cos \mathrm{\angle }A& =-4\\ -12\cos \mathrm{\angle }A-8\cos \mathrm{\angle }A& =-4\\ -20\cos \mathrm{\angle }A& =-4\\ \cos \mathrm{\angle }A& =\frac{-4}{-20}\\ \cos \mathrm{\angle }A& =\frac{1}{5}\end{array}\end{array}$.