Bank Soal: Aturan Sinus dan Pembahasan - CATATAN MATEMATIKA

Bank Soal: Aturan Sinus dan Pembahasan

Soal dan Pembahasan Aturan Sinus
Berikut ini adalah kumpulan Soal dan Pembahasan Aturan Sinus yaitu salah satu sub topik materi TRIGONOMETRI pada bidang studi Matematika Wajib kelas 10 Kurikulum 201. Selamat belajar ya...!
Soal No. 1
Pada segitiga ABC, jika diketahui a = 8 cm, b = $4\sqrt{2}$ cm, dan $\angle A=45{}^\circ $, maka $\angle B$ = …
(A) $30{}^\circ $
(B) $45{}^\circ $
(C) $55{}^\circ $
(D) $60{}^\circ $
(E) $78{}^\circ $
$\begin{align} \frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{4\sqrt{2}}{\sin B} &=\frac{8}{\sin 45{}^\circ } \\ \frac{4\sqrt{2}}{\sin B} &=\frac{8}{\frac{1}{2}\sqrt{2}} \\ 8\sin B &=4 \\ \sin B &=\frac{4}{8}=\frac{1}{2} \\ B &=30{}^\circ \end{align}$
Jawaban: A

Soal No. 2
Diketahui segitiga ABC dengan $\angle A=30{}^\circ $, $\angle C=105{}^\circ $, dan BC = 10 cm. Panjang AC = ….
(A) 5 cm
(B) $5\sqrt{3}$ cm
(C) $10\sqrt{2}$ cm
(D) $10\sqrt{3}$ cm
(E) $\frac{10}{3}\sqrt{3}$ cm
$\begin{align} \angle A+\angle B+\angle C &=180{}^\circ \\ 30{}^\circ +\angle B+105{}^\circ &=180{}^\circ \\ \angle B &=45{}^\circ \end{align}$
BC = 10 = a, AC = b = …?
$\begin{align}\frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{b}{\sin 45{}^\circ } &=\frac{10}{\sin 30{}^\circ } \\ \frac{b}{\frac{1}{2}\sqrt{2}} &=\frac{10}{\frac{1}{2}} \\ b &=10\sqrt{2} \end{align}$
Jawaban: C

Soal No. 3
Perhatikan gambar berikut!
Soal dan Pembahasan Aturan Sinus
BC : AC = ….
(A) 3 : 4
(B) 4 : 3
(C) $\sqrt{2}:\sqrt{3}$
(D) $\sqrt{3}:2\sqrt{2}$
(E) $\sqrt{3}:\sqrt{2}$
BC : AC = a : b = …?
$\begin{align} \frac{a}{\sin A} &=\frac{b}{\sin B} \\ \frac{a}{\sin 45{}^\circ } &=\frac{b}{\sin 60{}^\circ } \\ \frac{a}{b} &=\frac{\sin 45{}^\circ }{\sin 60{}^\circ } \\ \frac{a}{b} &=\frac{\frac{1}{2}\sqrt{2}}{\frac{1}{2}\sqrt{3}} \\ \frac{a}{b} &=\frac{\sqrt{2}}{\sqrt{3}} \end{align}$
Jawaban: C

Soal No. 4
Perhatikan gambar berikut!
Soal dan Pembahasan Aturan Sinus
Dua orang mulai berjalan masing-masing dari titik A dan titik B pada saat yang sama. Supaya keduanya sampai di titik C pada saat yang sama, maka kecepatan berjalan dari titik A harus ….
(A) 2 kali kecepatan orang yang dari B
(B) $\frac{1}{2}\sqrt{2}$ kali kecepatan orang yang dari B.
(C) $\sqrt{2}$ kali kecepatan orang yang dari B.
(D) $2\sqrt{2}$ kali kecepatan orang yang dari B.
(E) $\sqrt{3}$ kali kecepatan orang yang dari B.
Aturan sinus:
$\begin{align} \frac{a}{\sin A} &=\frac{b}{\sin B} \\ \frac{\sin B}{\sin A} &=\frac{b}{a} \end{align}$
$\begin{align} {{t}_{a}} &={{t}_{b}} \\ \frac{{{s}_{a}}}{{{v}_{a}}} &=\frac{{{s}_{b}}}{{{v}_{b}}} \\ \frac{b}{{{v}_{a}}} &=\frac{a}{{{v}_{b}}} \\ \frac{{{v}_{a}}}{{{v}_{b}}} &=\frac{b}{a} \\ \frac{{{v}_{a}}}{{{v}_{b}}} &=\frac{\sin B}{\sin A} \\ \frac{{{v}_{a}}}{{{v}_{b}}} &=\frac{\sin 30{}^\circ }{\sin 45{}^\circ } \\ \frac{{{v}_{a}}}{{{v}_{b}}} &=\frac{\frac{1}{2}}{\frac{1}{2}\sqrt{2}} \\ {{v}_{a}} &=\frac{1}{\sqrt{2}}.{{v}_{b}} \\ {{v}_{a}} &=\frac{1}{2}\sqrt{2}.{{v}_{b}} \end{align}$
Jawaban: A

Soal No. 5
Diketahui segitiga PQR dengan PR = 3 cm dan QR = $\frac{3\sqrt{6}}{2}$, $\angle P=60{}^\circ $. Besar sudut R adalah ….
(A) $30{}^\circ $
(B) $45{}^\circ $
(C) $75{}^\circ $
(D) $105{}^\circ $
(E) $120{}^\circ $
PR = q = 3 cm, QR = p = $\frac{3\sqrt{6}}{2}$, $\angle P=60{}^\circ $
$\begin{align} \frac{q}{\sin Q} &=\frac{p}{\sin P} \\ \frac{3}{\sin Q} &=\frac{\frac{3\sqrt{6}}{2}}{\sin 60{}^\circ } \\ \frac{3}{\sin Q} &=\frac{\frac{3}{2}\sqrt{6}}{\frac{1}{2}\sqrt{3}} \\ \frac{3}{\sin Q} &=3\sqrt{2} \\ 3\sqrt{2}.\sin Q &=3 \\ \sin Q &=\frac{1}{\sqrt{2}} \\ \sin Q &=\frac{1}{2}\sqrt{2} \\ Q &=45{}^\circ \end{align}$
$\begin{align} P+Q+R &=180{}^\circ \\ 60{}^\circ +45{}^\circ +R & =180{}^\circ \\ 105{}^\circ +R &=180{}^\circ \\ R &=180{}^\circ -105{}^\circ \\ R &=75{}^\circ \end{align}$
Jawaban: C
Soal No. 6
Suatu segitiga ABC dan diketahui $\angle A=45{}^\circ $, AC = 2 cm, BC = $2\sqrt{2}$ maka nilai $\cos \angle B$ = …
(A) $\frac{1}{2}$
(B) $\frac{1}{2}\sqrt{2}$
(C) $\frac{1}{2}\sqrt{3}$
(D) $\frac{1}{2}\sqrt{5}$
(E) 1
$\angle A=45{}^\circ $, AC = b = 2 cm, BC = a = $2\sqrt{2}$
$\begin{align} \frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{2}{\sin B} &=\frac{2\sqrt{2}}{\sin 45{}^\circ } \\ \frac{1}{\sin B} &=\frac{\sqrt{2}}{\frac{1}{2}\sqrt{2}} \\ \frac{1}{\sin B} &=\frac{2}{1} \\ \sin B &=\frac{1}{2} \\ B &=30{}^\circ \end{align}$
$\cos B =\cos 30{}^\circ =\frac{1}{2}\sqrt{3}$
Jawaban: C

Soal No. 7
Perhatikan gambar berikut!
Soal dan Pembahasan Aturan Sinus
Nilai dari sin D = ….
(A) $\frac{1}{16}\sqrt{6}$
(B) $\frac{1}{8}\sqrt{6}$
(C) $\frac{1}{4}\sqrt{6}$
(D) $\frac{1}{3}\sqrt{6}$
(E) $\frac{1}{5}\sqrt{6}$
Perhatikan segitiga ABC:
$\begin{align} \frac{BC}{\sin \angle BAC} &=\frac{AC}{\sin \angle ABC} \\ \frac{BC}{\sin 30{}^\circ } &=\frac{2}{\sin 45{}^\circ } \\ \frac{BC}{\frac{1}{2}} &=\frac{2}{\frac{1}{2}\sqrt{2}} \\ BC &=\frac{2}{\sqrt{2}} \\ BC &=\sqrt{2} \end{align}$
Perhatikan segitiga BCD:
$\begin{align} \frac{BC}{\sin \angle BDC} &=\frac{BD}{\sin \angle BCD} \\ \frac{\sqrt{2}}{\sin D} &=\frac{4}{\sin 60{}^\circ } \\ \frac{\sqrt{2}}{\sin D} &=\frac{4}{\frac{1}{2}\sqrt{3}} \\ 4\sin D &=\frac{1}{2}\sqrt{6} \\ \sin D &=\frac{1}{8}\sqrt{6} \end{align}$
Jawaban: B

Soal No. 8
Pada segitiga ABC diketahui sisi $a=4$, sisi $b=6$ dan sudut B = $45{}^\circ $. Nilai kosinus sudut A adalah …..
(A) $\frac{1}{6}\sqrt{2}$
(B) $\frac{1}{6}\sqrt{6}$
(C) $\frac{1}{6}\sqrt{7}$
(D) $\frac{1}{3}\sqrt{2}$
(E) $\frac{1}{3}\sqrt{7}$
$a=4$, sisi $b=6$, $\angle B$ = $45{}^\circ $, $\cos A$ = …?
$\begin{align} \frac{a}{\sin A} &=\frac{b}{\sin B} \\ \frac{4}{\sin A} &=\frac{6}{\sin 45{}^\circ } \\ \frac{4}{\sin A} &=\frac{6}{\frac{1}{2}\sqrt{2}} \\ 6\sin A &=2\sqrt{2} \\ \sin A &=\frac{\sqrt{2}}{3} \end{align}$
$\sin A=\frac{de}{mi}=\frac{\sqrt{2}}{3}$
$\begin{align} sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ & =\sqrt{{{3}^{2}}-{{(\sqrt{2})}^{2}}} \\ sa &=\sqrt{7} \end{align}$
$\cos A=\frac{sa}{mi}=\frac{\sqrt{7}}{3}$
Jawaban: E

Soal No. 9
Diketahui segitiga PQR dengan panjang QR = $\sqrt{6}$, PR = $\sqrt{2}$ dan $\angle P=60{}^\circ $, maka panjang PQ = …
(A) 8
(B) 2
(C) $2\sqrt{6}$
(D) $2\sqrt{2}$
(E) $2\sqrt{10}$
QR = p = $\sqrt{6}$, PR = q = $\sqrt{2}$ dan $\angle P=60{}^\circ $
PQ = r = …
$\begin{align} \frac{q}{\sin Q} &=\frac{p}{\sin P} \\ \frac{\sqrt{2}}{\sin Q} &=\frac{\sqrt{6}}{\sin 60{}^\circ } \\ \frac{\sqrt{2}}{\sin Q} &=\frac{\sqrt{6}}{\frac{1}{2}\sqrt{3}} \\ \sqrt{6}.\sin Q &=\frac{1}{2}\sqrt{6} \\ \sin Q &=\frac{1}{2} \\ Q &=30{}^\circ \end{align}$
$\begin{align} P+Q+R &=180{}^\circ \\ 60{}^\circ +30{}^\circ +R &=180{}^\circ \\ R &=90{}^\circ \end{align}$
$\begin{align} \frac{r}{\sin R} &=\frac{q}{\sin Q} \\ \frac{r}{\sin 90{}^\circ } &=\frac{\sqrt{2}}{\sin 30{}^\circ } \\ \frac{r}{1} &=\frac{\sqrt{2}}{\frac{1}{2}} \\ r &=2\sqrt{2} \end{align}$
Jawaban: D

Soal No. 10
Diketahui segitiga ABC dengan panjang sisi a = $3\sqrt{2}$ cm, b = 6 cm, dan $\angle A=30{}^\circ $ maka nilai $\tan \angle B$ = …
(A) $\frac{1}{2}\sqrt{2}$
(B) $\frac{1}{2}\sqrt{3}$
(C) 1
(D) $-1$
(E) 0
$\begin{align} \frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{6}{\sin B} &=\frac{3\sqrt{2}}{\sin 30{}^\circ } \\ \frac{6}{\sin B} &=\frac{3\sqrt{2}}{\frac{1}{2}} \\ 3\sqrt{2}.\sin B &=3 \\ \sin B &=\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2} \\ B &=45{}^\circ \end{align}$
$\tan B=\tan 45{}^\circ =1$
Jawaban: C
Soal No. 11
Dalam segitiga ABC diketahui a = 9, $\angle A=60{}^\circ $, $\angle B=45{}^\circ $ maka b = ….
(A) $9\sqrt{3}$
(B) $\frac{9}{2}\sqrt{3}$
(C) $3\sqrt{6}$
(D) $\frac{1}{2}\sqrt{6}$
(E) $3\sqrt{3}$
$\begin{align} \frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{b}{\sin 45{}^\circ } &=\frac{9}{\sin 60{}^\circ } \\ \frac{b}{\frac{1}{2}\sqrt{2}} &=\frac{9}{\frac{1}{2}\sqrt{3}} \\ b &=\frac{9\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ b &=3\sqrt{6} \end{align}$
Jawaban: C

Soal No. 12
Suatu segitiga PQR dengan sisi p = 9 cm, $\angle P=120{}^\circ $, $\angle Q=45{}^\circ $, maka panjang sisi q = … cm.
(A) $3\sqrt{6}$
(B) $9\sqrt{2}$
(C) $\frac{9}{2}\sqrt{6}$
(D) $9\sqrt{6}$
(E) $6\sqrt{6}$
$\begin{align} \frac{q}{\sin Q} &=\frac{p}{\sin P} \\ \frac{q}{\sin 45{}^\circ } &=\frac{9}{\sin 120{}^\circ } \\ \frac{q}{\frac{1}{2}\sqrt{2}} &=\frac{9}{\sin (180{}^\circ -60{}^\circ )} \\ q &=\frac{\frac{9\sqrt{2}}{2}}{\sin 60{}^\circ } \\ q &=\frac{\frac{9\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} \\ q &=\frac{9\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ q &=3\sqrt{6} \end{align}$
Jawaban: A

Soal No. 13
Diketahui segitiga PQR dimana p = 10 cm, $\angle P=45{}^\circ $, dan $\angle Q=105{}^\circ $, panjang sisi r = … cm.
(A) 5
(B) $10\sqrt{2}$
(C) $5\sqrt{2}$
(D) 10
(E) 20
$\begin{align} P+Q+R &=180{}^\circ \\ 45{}^\circ +105{}^\circ +R &=180{}^\circ \\ R &=30{}^\circ \end{align}$
$\begin{align} \frac{r}{\sin R} &=\frac{p}{\sin P} \\ \frac{r}{\sin 30{}^\circ } &=\frac{10}{\sin 45{}^\circ } \\ \frac{r}{\frac{1}{2}} &=\frac{10}{\frac{1}{2}\sqrt{2}} \\ r &=\frac{10}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ r &=5\sqrt{2} \end{align}$
Jawaban: C

Soal No. 14
Diketahui segitiga ABC dengan besar $\angle A=60{}^\circ $, $\angle B=75{}^\circ $, dan panjang AB = 12 cm. Panjang BC adalah … cm.
(A) $6\sqrt{2}$
(B) $6\sqrt{3}$
(C) $6\sqrt{6}$
(D) $12\sqrt{3}$
(E) $12\sqrt{6}$
$\angle A=60{}^\circ $, $\angle B=75{}^\circ $, AB = c = 12 cm,
BC = a = …?
$\begin{align} A+B+C &=180{}^\circ \\ 60{}^\circ +75{}^\circ +C &=180{}^\circ \\ C &=45{}^\circ \end{align}$
$\begin{align} \frac{a}{\sin A} &=\frac{c}{\sin C} \\ \frac{a}{\sin 60{}^\circ } &=\frac{12}{\sin 45{}^\circ } \\ \frac{a}{\frac{1}{2}\sqrt{3}} &=\frac{12}{\frac{1}{2}\sqrt{2}} \\ a &=\frac{12\sqrt{3}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ a &=6\sqrt{6} \\ BC &=6\sqrt{6} \end{align}$
Jawaban: C

Soal No. 15
Pada segitiga RST diketahui ST = 4 cm, $\angle R=30{}^\circ $ dan $\angle T=105{}^\circ $. Panjang sisi RT adalah … cm.
(A) $8\sqrt{2}$
(B) 6
(C) $5\sqrt{2}$
(D) $4\sqrt{2}$
(E) $3\sqrt{2}$
ST = r = 4 cm, $\angle R=30{}^\circ $, $\angle T=105{}^\circ $
RT = s = …?
$\begin{align} R+S+T &=180{}^\circ \\ 30{}^\circ +S+105{}^\circ &=180{}^\circ \\ S &=45{}^\circ \end{align}$
$\begin{align} \frac{s}{\sin S} &=\frac{r}{\sin R} \\ \frac{s}{\sin 45{}^\circ } &=\frac{4}{\sin 30{}^\circ } \\ \frac{s}{\frac{1}{2}\sqrt{2}} &=\frac{4}{\frac{1}{2}} \\ s &=4\sqrt{2} \end{align}$
Jawaban: D

Soal No. 16
Ditentukan segitiga ABC dengan panjang sisi BC = 3 cm, sisi AC = 4 cm dan $\sin A=\frac{1}{2}$. Nilai cos B = …
(A) $\frac{2}{5}\sqrt{5}$
(B) $\frac{1}{3}\sqrt{5}$
(C) $\frac{1}{2}\sqrt{3}$
(D) $\frac{2}{3}$
(E) $\frac{1}{2}$
BC = a = 3 cm, AC = b = 4 cm dan $\sin A=\frac{1}{2}$
$\begin{align} \frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{4}{\sin B} &=\frac{3}{\frac{1}{2}} \\ 3\sin B &=2 \\ \sin B &=\frac{2}{3} \end{align}$
$\sin B=\frac{de}{mi}=\frac{2}{3}$
$\begin{align}sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ &=\sqrt{{{3}^{2}}-{{2}^{2}}} \\ sa &=\sqrt{5} \end{align}$
$\cos B=\frac{sa}{mi}=\frac{\sqrt{5}}{3}$
Jawaban: B

Soal No. 17
Pada segitiga ABC. Diketahui $\angle C=105{}^\circ $, $\angle A=30{}^\circ $ dan BC = 4. Panjang AC = ….
(A) $2\sqrt{2}$
(B) $3\sqrt{2}$
(C) $4\sqrt{2}$
(D) 6
(E) 8
$\begin{align} \angle A+\angle B+\angle C &=180{}^\circ \\ 30{}^\circ +\angle B+105{}^\circ &=180{}^\circ \\ \angle B+135{}^\circ =180{}^\circ \\ \angle B &=45{}^\circ \end{align}$
BC = 4 = a, AC = b = …?
$\begin{align} \frac{b}{\sin B} &=\frac{a}{\sin A} \\ \frac{b}{\sin 45{}^\circ } &=\frac{4}{\sin 30{}^\circ } \\ \frac{b}{\frac{1}{2}\sqrt{2}} &=\frac{4}{\frac{1}{2}} \\ b &=4\sqrt{2} \end{align}$
Jawaban: C
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