Bank Soal Matematika SMA Aturan Sinus

Diketahui:

segitiga $KLM$

$KL=100$ cm,

$\angle MKL=60^o$

$\angle KLM=75^o$.

Ditanya:

Panjang $KM=?$

Jawab:

Ingat aturan sinus, jika diberikan segitiga sembarang $ABC$, maka berlaku:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

dengan $R$ adalah jari-jari lingkaran luar segitiga $ABC.$

Perhatikan gambar berikut!

Karena jumlah semua sudut segitiga adalah 180^o, maka

$\angle KML=180^o-75^o-60^o=45^o$

$\sin75^o=\sin\left(30^o+45^o\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ =\sin30^o\cos45^o+\cos30^o\sin45^o$

$\ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\cdot\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}\cdot\frac{1}{2}\sqrt{2}$

$\ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{3}\cdot\sqrt{2}$

$\ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{6}$

$\ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{4}\left(\sqrt{2}+\sqrt{6}\right)$

Dengan menggunakan aturan sinus, diperoleh

$\frac{KM}{\sin L}=\frac{KL}{\sin M}$

$\frac{KM}{\sin75^o}=\frac{100}{\sin45^o}$

$\frac{KM}{\frac{1}{4}\left(\sqrt{2}+\sqrt{6}\right)}=\frac{100}{\frac{1}{2}\sqrt{2}}$

$\frac{KM}{\frac{1}{2}\left(\sqrt{2}+\sqrt{6}\right)}=\frac{100}{\sqrt{2}}$

$KM=\frac{100}{\sqrt{2}}\cdot\frac{1}{2}\left(\sqrt{2}+\sqrt{6}\right)$

$KM=\frac{50}{\sqrt{2}}\cdot\left(\sqrt{2}+\sqrt{6}\right)$

$KM=\frac{50}{\sqrt{2}}\sqrt{2}+\frac{50}{\sqrt{2}}\sqrt{6}$

$KM=50+\frac{50}{\sqrt{2}}\sqrt{6}\cdot\frac{\sqrt{2}}{\sqrt{2}}$

$KM=50+\frac{50}{2}\sqrt{12}$

$KM=50+25\sqrt{4\cdot3}$

$KM=50+25\sqrt{4}\cdot\sqrt{3}$

$KM=50+25\cdot2\cdot\sqrt{3}$

$KM=50+50\sqrt{3}$

$KM=50\left(1+\sqrt{3}\right)$

Jadi, panjang $KM=50\left(1+\sqrt{3}\right)$ cm.