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Bank Soal: Perbandingan Trigonometri dan Pembahasan

Soal dan Pembahasan Perbandingan Trigonometri
Berikut ini adalah Soal dan Pembahasan Perbandingan Trigonometri yaitu salah satu sub materi TRIGONOMETRI bidang studi Matematika Wajib Kelas 10.
Soal No. 1
Pada segitiga PQR di bawah ini, $\sin \beta $ = …
Soal Perbandingan Trigonometri dan Pembahasan
(A) $\frac{p}{q}$
(B) $\frac{p}{r}$
(C) $\frac{r}{q}$
(D) $\frac{q}{p}$
(E) $\frac{r}{p}$
$\sin \beta =\frac{\text{sisi}\,\text{depan}}{\text{sisi}\,\text{miring}}=\frac{q}{p}$
Jawaban: D

Soal No. 2
Pada segitiga KLM di bawah ini nilai dari $\sin \alpha +\sin \beta $ = …
Soal Perbandingan Trigonometri dan Pembahasan
(A) $\frac{12}{10}$
(B) $\frac{14}{10}$
(C) $\frac{16}{10}$
(D) $\frac{18}{10}$
(E) $\frac{20}{10}$
$\begin{align} \sin \alpha +\sin \beta &=\frac{ML}{KL}+\frac{KM}{KL} \\ &=\frac{8}{10}+\frac{6}{10} \\ \sin \alpha +\sin \beta &=\frac{14}{10} \end{align}$
Jawaban: B

Soal No. 3
Jika $\sin \alpha =\frac{12}{13}$, dengan $\alpha $ lancip maka $\cos \alpha $ = ….
(A) $\frac{13}{12}$
(B) $\frac{13}{5}$
(C) $\frac{5}{13}$
(D) $\frac{12}{5}$
(E) $\frac{5}{12}$
$\sin \alpha =\frac{12}{13}=\frac{de}{mi}$
$\begin{align} sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ &=\sqrt{{{13}^{2}}-{{12}^{2}}} \\ &=\sqrt{169-144} \\ &=\sqrt{25} \\ sa &=5 \end{align}$
$\cos \alpha =\frac{sa}{mi}=\frac{5}{13}$
Jawaban: C

Soal No. 4
Jika $\cos A=\frac{2}{3}$, dengan A lancip maka $\tan A$ = …
(A) $\frac{1}{3}\sqrt{5}$
(B) $\frac{3}{2}$
(C) $\frac{2}{5}\sqrt{5}$
(D) $\frac{1}{2}\sqrt{5}$
(E) $\frac{3}{5}\sqrt{5}$
$\cos A=\frac{2}{3}=\frac{sa}{mi}$
$\begin{align} de &=\sqrt{m{{i}^{2}}-s{{a}^{2}}} \\ &=\sqrt{{{3}^{2}}-{{2}^{2}}} \\ &=\sqrt{9-4} \\ de &=\sqrt{5} \end{align}$
$\tan A=\frac{de}{sa}=\frac{\sqrt{5}}{2}$
Jawaban: D

Soal No. 5
Jika $\tan A=3$, dengan A lancip maka $\sin A$ = ….
(A) $\frac{1}{3}\sqrt{10}$
(B) $\frac{8}{10}\sqrt{10}$
(C) $\frac{10}{3}\sqrt{10}$
(D) $\frac{3}{10}\sqrt{10}$
(E) $\frac{1}{10}\sqrt{10}$
$\tan A=3=\frac{3}{1}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{3}^{2}}+{{1}^{2}}} \\ &=\sqrt{9+1} \\ mi &=\sqrt{10} \end{align}$
$\begin{align} \sin A &=\frac{de}{mi} \\ &=\frac{3}{\sqrt{10}} \\ &=\frac{3}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}} \\ \sin A &=\frac{3}{10}\sqrt{10} \end{align}$
Jawaban: D
Soal No. 6
Bila $0{}^\circ < a < 90{}^\circ $ dan $\tan a=\frac{5}{\sqrt{11}}$, maka $\sin a$ = …
(A) $\frac{5}{6}$
(B) $\frac{25}{36}$
(C) $\frac{1}{6}\sqrt{11}$
(D) $\frac{5}{36}$
(E) $\frac{1}{36}\sqrt{11}$
$\tan a=\frac{5}{\sqrt{11}}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{5}^{2}}+{{(\sqrt{11})}^{2}}} \\ &=\sqrt{25+11} \\ &=\sqrt{36} \\ mi &=6 \end{align}$
$\sin a=\frac{de}{mi}=\frac{5}{6}$
Jawaban: A

Soal No. 7
Pada gambar disamping nilai $\cos \angle BAC$ adalah ….
Soal Perbandingan Trigonometri dan Pembahasan
(A) $\frac{15}{40}$
(B) $\frac{15}{25}$
(C) $\frac{15}{20}$
(D) $\frac{20}{25}$
(E) $\frac{25}{40}$
$\begin{align} AC &=\sqrt{A{{B}^{2}}-B{{C}^{2}}} \\ & =\sqrt{{{25}^{2}}-{{15}^{2}}} \\ & =\sqrt{625-225} \\ & =\sqrt{400} \\ AC &=20 \end{align}$
$\cos \angle BAC=\frac{AC}{AB}=\frac{20}{25}$
Jawaban: D

Soal No. 8
Jika $\cos x=2\sin x$, nilai $\sin x\cos x$ adalah …
(A) $\frac{1}{5}$
(B) $\frac{1}{4}$
(C) $\frac{1}{3}$
(D) $\frac{2}{5}$
(E) $\frac{2}{3}$
$\begin{align} \cos x &=2\sin x \\ \frac{1}{2} &=\frac{\sin x}{\cos x} \\ \tan x=\frac{1}{2} &=\frac{de}{sa} \end{align}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ & =\sqrt{{{1}^{2}}+{{2}^{3}}} \\ mi &=\sqrt{5} \end{align}$
$\begin{align} \sin x\cos x &=\frac{de}{mi}\times \frac{sa}{mi} \\ & =\frac{1}{\sqrt{5}}\times \frac{2}{\sqrt{5}} \\ \sin x.\cos x &=\frac{2}{5} \end{align}$
Jawaban: D

Soal No. 9
Apabila $4\cot x=3$ dengan sudut $x$ lancip maka nilai dari $\frac{\sin x-\cos x}{\sin x+\cos x}$ = ….
(A) $-\frac{1}{7}$
(B) 0
(C) $\frac{1}{7}$
(D) $\frac{7}{3}$
(E) $\frac{3}{7}$
$4\cot x=3\Leftrightarrow \cot x=\frac{3}{4}$
$\begin{align}\frac{\sin x-\cos x}{\sin x+\cos x} &=\frac{\frac{\sin x}{\sin x}-\frac{\cos x}{\sin x}}{\frac{\sin x}{\sin x}+\frac{\cos x}{\sin x}} \\ & =\frac{1-\cot x}{1+\cot x} \\ & =\frac{1-\frac{3}{4}}{1+\frac{3}{4}} \\ \frac{\sin x-\cos x}{\sin x+\cos x} &=\frac{1}{7} \end{align}$
Jawaban: C

Soal No. 10
Perhatikan gambar berikut!
Soal Perbandingan Trigonometri dan Pembahasan
Diketahui $\Delta ABC$ siku-siku di B, $\cos \alpha =\frac{12}{13}$, dan $\tan \beta =1$. Jika$AD=a$, maka $AC$ = …
(A) $\frac{1}{2}a$
(B) $\frac{11}{7}a$
(C) $\frac{12}{7}a$
(D) $\frac{13}{7}a$
(E) $2a$
Perhatikan segitiga CBD:
$\begin{align} \cos \alpha &=\frac{12}{13} \\ \frac{AB}{AC} &=\frac{12}{13} \end{align}$
$\begin{align} BC &=\sqrt{A{{C}^{2}}-A{{B}^{2}}} \\ &=\sqrt{{{13}^{2}}-{{12}^{2}}} \\ BC &=5 \end{align}$
$\begin{align} \tan \beta &=1 \\ \frac{BC}{BD} &=1 \\ BC&=BD=5 \end{align}$
$\begin{align} AB &=AD+BD \\ 12 &=a+5 \\ a &=7 \end{align}$
$\begin{align} AC &=13 \\ &=\frac{13}{7}\times 7 \\ AC &=\frac{13}{7}a \end{align}$
Jawaban: D
Soal No. 11
Jika $\sin x=k$, maka $\frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ = ….
(A) $1+k$
(B) $1+{{k}^{2}}$
(C) $1+2{{k}^{2}}$
(D) $1-2{{k}^{2}}$
(E) 1
$\begin{align} \sin x &=k \\ \frac{de}{mi} &=\frac{k}{1} \\ sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ sa &=\sqrt{1-{{k}^{2}}} \end{align}$
$\tan x=\frac{de}{sa}=\frac{k}{\sqrt{1-{{k}^{2}}}}$
${{\tan }^{2}}x=\frac{{{k}^{2}}}{1-{{k}^{2}}}$
$\begin{align} \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} &=\frac{1-\frac{{{k}^{2}}}{1-{{k}^{2}}}}{1-\frac{{{k}^{2}}}{1-{{k}^{2}}}} \\ &=\frac{\frac{1-{{k}^{2}}-{{k}^{2}}}{1-{{k}^{2}}}}{\frac{1-{{k}^{2}}+{{k}^{2}}}{1-{{k}^{2}}}} \\ \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} &=1-2{{k}^{2}} \end{align}$
Jawaban: D

Soal No. 12
Jika $\tan \alpha =-p$ dengan $\alpha $ sudut lancip maka $\sin \alpha $ = …
(A) $\frac{-2p}{\sqrt{{{p}^{2}}+1}}$
(B) $\frac{-p}{\sqrt{{{p}^{2}}+1}}$
(C) $\frac{p}{\sqrt{{{p}^{2}}+1}}$
(D) $\frac{2p}{\sqrt{{{p}^{2}}+1}}$
(E) $\frac{1}{\sqrt{{{p}^{2}}+1}}$
$\begin{align} \tan \alpha &=-p \\ \frac{de}{sa} &=\frac{-p}{1} \\ mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{(-p)}^{2}}+{{1}^{2}}} \\ mi &=\sqrt{{{p}^{2}}+1} \end{align}$
$\sin \alpha =\frac{de}{mi}=\frac{-p}{\sqrt{{{p}^{2}}+1}}$
Jawaban: B

Soal No. 13
Jika $0 < x < \frac{\pi }{2}$ dan $x$ memenuhi persamaan $6{{\sin }^{2}}x-\sin x-1=0$, maka $\cos x$ = ….
(A) $\frac{2\sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{2}}{3}$
(D) $\frac{\sqrt{3}}{3}$
(E) $\frac{\sqrt{3}}{4}$
Misal:
$\sin x=p$ maka:
$\begin{align} 6{{\sin }^{2}}x-\sin x-1 &=0 \\ 6{{p}^{2}}-p-1 &=0 \\ (3p+1)(2p-1) &=0 \end{align}$
$\begin{align} p &=-\frac{1}{3} \\ \sin x &=-\frac{1}{3} \end{align}$ atau $\begin{align} p &=\frac{1}{2} \\ \sin x &=\frac{1}{2} \end{align}$
Karena $0 < x < \frac{\pi }{2}$ maka yang memenuhi adalah:
$\begin{align} \sin x &=\frac{1}{2} \\ \frac{de}{mi} &=\frac{1}{2} \\ sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ &=\sqrt{{{2}^{2}}-{{1}^{2}}} \\ sa &=\sqrt{3} \\ \cos x &=\frac{sa}{mi}=\frac{\sqrt{3}}{2} \end{align}$
Jawaban: B

Soal No. 14
Apabila $\tan \theta =\frac{1}{\sqrt{7}}$ maka nilai dari $\frac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }$ adalah ….
(A) $\frac{1}{12}$
(B) $\frac{3}{7}$
(C) $\frac{3}{4}$
(D) $\frac{5}{7}$
(E) $\frac{4}{3}$
$\begin{align} \frac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta } &=\frac{\frac{1}{{{\sin }^{2}}\theta }-\frac{1}{{{\cos }^{2}}\theta }}{\frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta }} \\ &=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \\ &=\frac{1-{{\left( \frac{1}{\sqrt{7}} \right)}^{2}}}{1+{{\left( \frac{1}{\sqrt{7}} \right)}^{2}}} \\ &=\frac{1-\frac{1}{7}}{1+\frac{1}{7}} \\ &=\frac{6}{8} \\ \frac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta } &=\frac{3}{4} \end{align}$
Jawaban: C

Soal No. 15
Apabila $\tan A=t$ dengan A sudut lancip, maka $\sin A$ = ….
(A) $1+{{t}^{2}}$
(B) $1-{{t}^{2}}$
(C) $\sqrt{\frac{1}{{{t}^{2}}+1}}$
(D) $\sqrt{{{t}^{2}}+1}$
(E) $\sqrt{\frac{{{t}^{2}}}{{{t}^{2}}+1}}$
$\begin{align} \tan A &=t \\ \frac{de}{sa} &=\frac{t}{1} \\ mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}}=\sqrt{{{t}^{2}}+1} \\ \sin A &=\frac{de}{mi} \\ &=\frac{t}{\sqrt{{{t}^{2}}+1}} \\ \sin A &=\sqrt{\frac{{{t}^{2}}}{{{t}^{2}}+1}} \end{align}$
Jawaban: E
Soal No. 16
Pada segitiga ABC siku-siku di B dengan $\cos C=\frac{1}{3}$. Jika BD adalah garis tinggi pada sisi AC dan BD = 6 cm, maka panjang AC = … cm.
(A) $4\sqrt{2}$
(B) $\frac{27}{2}\sqrt{2}$
(C) $5\sqrt{2}$
(D) $\frac{11}{2}\sqrt{2}$
(E) $6\sqrt{2}$
  Soal Perbandingan Trigonometri dan Pembahasan
Perhatikan segitiga BDC:
$\begin{align} \cos C &=\frac{CD}{BC} \\ \frac{1}{3} &=\frac{CD}{BC} \\ BC &=3CD \end{align}$
Teorema pythagoras:
$\begin{align} B{{C}^{2}} &=B{{D}^{2}}+C{{D}^{2}} \\ {{(3CD)}^{2}} &={{6}^{2}}+C{{D}^{2}} \\ 9C{{D}^{2}} &=36+C{{D}^{2}} \\ 8C{{D}^{2}} &=36 \\ C{{D}^{2}} &=\frac{36}{8}=\frac{9}{2} \\ CD &=\frac{3}{\sqrt{2}} \end{align}$
$\begin{align} BC &=3CD \\ &=3.\frac{3}{\sqrt{2}} \\ BC &=\frac{9}{\sqrt{2}} \end{align}$
Perhatikan segitiga ABC:
$\begin{align} \cos C &=\frac{BC}{AC} \\ \frac{1}{3} &=\frac{\frac{9}{\sqrt{2}}}{AC} \\ AC &=\frac{27}{\sqrt{2}}=\frac{27}{2}\sqrt{2} \end{align}$
Jawaban: B

Soal No. 17
Perhatikan gambar berikut!
Soal Perbandingan Trigonometri dan Pembahasan
Nilai $\sin \theta $ dari segitiga di atas adalah …
(A) $\frac{b}{c}$
(B) $\frac{a}{c}$
(C) $\frac{b}{a}$
(D) $\frac{c}{d}$
(E) $\frac{c}{a}$
$\sin \alpha =\frac{\text{sisi}\,\text{depan}}{\text{hipotenusa}}=\frac{a}{c}$
Jawaban: B

Soal No. 18
Diketahui sin A = $\frac{p}{q}$ dengan $0{}^\circ < A < 90{}^\circ $ maka nilai dari ${{\cos }^{2}}A$ = ….
(A) $\frac{{{p}^{2}}}{{{p}^{2}}+{{q}^{2}}}$
(B) $\frac{{{p}^{2}}}{{{p}^{2}}-{{q}^{2}}}$
(C) $\frac{{{q}^{2}}}{{{p}^{2}}-{{q}^{2}}}$
(D) $\frac{{{q}^{2}}+{{p}^{2}}}{{{q}^{2}}}$
(E) $\frac{{{q}^{2}}-{{p}^{2}}}{{{q}^{2}}}$
$\sin A=\frac{p}{q}=\frac{de}{mi}$
$sa=\sqrt{m{{i}^{2}}-d{{e}^{2}}}=\sqrt{{{q}^{2}}-{{p}^{2}}}$
$\begin{align} {{\cos }^{2}}A &={{\left( \frac{sa}{mi} \right)}^{2}} \\ &={{\left( \frac{\sqrt{{{q}^{2}}-{{p}^{2}}}}{q} \right)}^{2}} \\ {{\cos }^{2}}A &=\frac{{{q}^{2}}-{{p}^{2}}}{{{q}^{2}}} \end{align}$
Jawaban: E

Soal No. 19
Diketahui nilai $\tan \alpha =\frac{1}{3}$ untuk $0{}^\circ < \alpha < 90{}^\circ $. Nilai dari $2\sin \alpha .\cos \alpha $ = …
(A) $\frac{3}{5}$
(B) $\frac{2}{5}$
(C) $\frac{1}{5}$
(D) $-\frac{2}{5}$
(E) $-\frac{3}{5}$
$\tan \alpha =\frac{1}{3}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{1}^{2}}+{{3}^{2}}} \\ mi &=\sqrt{10} \end{align}$
$\begin{align} 2\sin \alpha .\cos \alpha &=2.\frac{de}{mi}.\frac{sa}{mi} \\ &=2.\frac{1}{\sqrt{10}}.\frac{3}{\sqrt{10}} \\ &=\frac{6}{10} \\ 2\sin \alpha .\cos \alpha &=\frac{3}{5} \end{align}$
Jawaban: A

Soal No. 20
Diketahui nilai dari $\sin 25{}^\circ =p$. Nilai dari $\tan 25{}^\circ $ = …
(A) $\sqrt{1-{{p}^{2}}}$
(B) $\sqrt{1+{{p}^{2}}}$
(C) $\frac{p}{\sqrt{1-{{p}^{2}}}}$
(D) $\frac{p}{\sqrt{1+{{p}^{2}}}}$
(E) $\frac{2p}{\sqrt{1+{{p}^{2}}}}$
$\sin 25{}^\circ =p\Leftrightarrow \frac{de}{mi}=\frac{p}{1}$
$sa=\sqrt{m{{i}^{2}}-d{{e}^{2}}}=\sqrt{1-{{p}^{2}}}$
$\tan 25{}^\circ =\frac{de}{sa}=\frac{p}{\sqrt{1-{{p}^{2}}}}$
Jawaban: C

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