Soal Perbandingan Trigonometri pada Segitiga Siku-siku dan Pembahasan

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 $ = …

(A) $\frac{p}{q}$
(B) $\frac{p}{r}$
(C) $\frac{r}{q}$
(D) $\frac{q}{p}$
(E) $\frac{r}{p}$

Penyelesaian: Lihat/Tutup

$\sin \beta =\frac{\text{sisi}\,\text{depan}}{\text{sisi}\,\text{miring}}=\frac{q}{p}$
Jawaban: D

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Soal No. 2

Pada segitiga KLM di bawah ini nilai dari $\sin \alpha +\sin \beta $ = …

(A) $\frac{12}{10}$
(B) $\frac{14}{10}$
(C) $\frac{16}{10}$
(D) $\frac{18}{10}$
(E) $\frac{20}{10}$

Penyelesaian: Lihat/Tutup

$\begin{align} \sin \alpha +\sin \beta &=\frac{ML}{KL}+\frac{KM}{KL} \\ &=\frac{8}{10}+\frac{6}{10} \\ \sin \alpha +\sin \beta &=\frac{14}{10} \end{align}$
Jawaban: B

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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) $\frac{5}{12}$

Penyelesaian: Lihat/Tutup

$\sin \alpha =\frac{12}{13}=\frac{de}{mi}$
$\begin{align} sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ &=\sqrt{{{13}^{2}}-{{12}^{2}}} \\ &=\sqrt{169-144} \\ &=\sqrt{25} \\ sa &=5 \end{align}$
$\cos \alpha =\frac{sa}{mi}=\frac{5}{13}$
Jawaban: C

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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) $\frac{3}{5}\sqrt{5}$

Penyelesaian: Lihat/Tutup

$\cos A=\frac{2}{3}=\frac{sa}{mi}$
$\begin{align} de &=\sqrt{m{{i}^{2}}-s{{a}^{2}}} \\ &=\sqrt{{{3}^{2}}-{{2}^{2}}} \\ &=\sqrt{9-4} \\ de &=\sqrt{5} \end{align}$
$\tan A=\frac{de}{sa}=\frac{\sqrt{5}}{2}$
Jawaban: D

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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) $\frac{1}{10}\sqrt{10}$

Penyelesaian: Lihat/Tutup

$\tan A=3=\frac{3}{1}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{3}^{2}}+{{1}^{2}}} \\ &=\sqrt{9+1} \\ mi &=\sqrt{10} \end{align}$
$\begin{align} \sin A &=\frac{de}{mi} \\ &=\frac{3}{\sqrt{10}} \\ &=\frac{3}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}} \\ \sin A &=\frac{3}{10}\sqrt{10} \end{align}$
Jawaban: D

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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) $\frac{1}{36}\sqrt{11}$

Penyelesaian: Lihat/Tutup

$\tan a=\frac{5}{\sqrt{11}}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{5}^{2}}+{{(\sqrt{11})}^{2}}} \\ &=\sqrt{25+11} \\ &=\sqrt{36} \\ mi &=6 \end{align}$
$\sin a=\frac{de}{mi}=\frac{5}{6}$
Jawaban: A

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Soal No. 7

Pada gambar disamping nilai $\cos \angle BAC$ adalah ….

(A) $\frac{15}{40}$
(B) $\frac{15}{25}$
(C) $\frac{15}{20}$
(D) $\frac{20}{25}$
(E) $\frac{25}{40}$

Penyelesaian: Lihat/Tutup

$\begin{align} AC &=\sqrt{A{{B}^{2}}-B{{C}^{2}}} \\ & =\sqrt{{{25}^{2}}-{{15}^{2}}} \\ & =\sqrt{625-225} \\ & =\sqrt{400} \\ AC &=20 \end{align}$
$\cos \angle BAC=\frac{AC}{AB}=\frac{20}{25}$
Jawaban: D

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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) $\frac{2}{3}$

Penyelesaian: Lihat/Tutup

$\begin{align} \cos x &=2\sin x \\ \frac{1}{2} &=\frac{\sin x}{\cos x} \\ \tan x=\frac{1}{2} &=\frac{de}{sa} \end{align}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ & =\sqrt{{{1}^{2}}+{{2}^{3}}} \\ mi &=\sqrt{5} \end{align}$
$\begin{align} \sin x\cos x &=\frac{de}{mi}\times \frac{sa}{mi} \\ & =\frac{1}{\sqrt{5}}\times \frac{2}{\sqrt{5}} \\ \sin x.\cos x &=\frac{2}{5} \end{align}$
Jawaban: D

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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) $\frac{3}{7}$

Penyelesaian: Lihat/Tutup

$4\cot x=3\Leftrightarrow \cot x=\frac{3}{4}$
$\begin{align}\frac{\sin x-\cos x}{\sin x+\cos x} &=\frac{\frac{\sin x}{\sin x}-\frac{\cos x}{\sin x}}{\frac{\sin x}{\sin x}+\frac{\cos x}{\sin x}} \\ & =\frac{1-\cot x}{1+\cot x} \\ & =\frac{1-\frac{3}{4}}{1+\frac{3}{4}} \\ \frac{\sin x-\cos x}{\sin x+\cos x} &=\frac{1}{7} \end{align}$
Jawaban: C

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Soal No. 10

Perhatikan gambar berikut!

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) $2a$

Penyelesaian: Lihat/Tutup

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

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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

Penyelesaian: Lihat/Tutup

$\begin{align} \sin x &=k \\ \frac{de}{mi} &=\frac{k}{1} \\ sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ sa &=\sqrt{1-{{k}^{2}}} \end{align}$
$\tan x=\frac{de}{sa}=\frac{k}{\sqrt{1-{{k}^{2}}}}$
${{\tan }^{2}}x=\frac{{{k}^{2}}}{1-{{k}^{2}}}$
$\begin{align} \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} &=\frac{1-\frac{{{k}^{2}}}{1-{{k}^{2}}}}{1-\frac{{{k}^{2}}}{1-{{k}^{2}}}} \\ &=\frac{\frac{1-{{k}^{2}}-{{k}^{2}}}{1-{{k}^{2}}}}{\frac{1-{{k}^{2}}+{{k}^{2}}}{1-{{k}^{2}}}} \\ \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} &=1-2{{k}^{2}} \end{align}$
Jawaban: D

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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) $\frac{1}{\sqrt{{{p}^{2}}+1}}$

Penyelesaian: Lihat/Tutup

$\begin{align} \tan \alpha &=-p \\ \frac{de}{sa} &=\frac{-p}{1} \\ mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{(-p)}^{2}}+{{1}^{2}}} \\ mi &=\sqrt{{{p}^{2}}+1} \end{align}$
$\sin \alpha =\frac{de}{mi}=\frac{-p}{\sqrt{{{p}^{2}}+1}}$
Jawaban: B

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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) $\frac{\sqrt{3}}{4}$

Penyelesaian: Lihat/Tutup

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

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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) $\frac{4}{3}$

Penyelesaian: Lihat/Tutup

$\begin{align} \frac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta } &=\frac{\frac{1}{{{\sin }^{2}}\theta }-\frac{1}{{{\cos }^{2}}\theta }}{\frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta }} \\ &=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \\ &=\frac{1-{{\left( \frac{1}{\sqrt{7}} \right)}^{2}}}{1+{{\left( \frac{1}{\sqrt{7}} \right)}^{2}}} \\ &=\frac{1-\frac{1}{7}}{1+\frac{1}{7}} \\ &=\frac{6}{8} \\ \frac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta } &=\frac{3}{4} \end{align}$
Jawaban: C

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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) $\sqrt{\frac{{{t}^{2}}}{{{t}^{2}}+1}}$

Penyelesaian: Lihat/Tutup

$\begin{align} \tan A &=t \\ \frac{de}{sa} &=\frac{t}{1} \\ mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}}=\sqrt{{{t}^{2}}+1} \\ \sin A &=\frac{de}{mi} \\ &=\frac{t}{\sqrt{{{t}^{2}}+1}} \\ \sin A &=\sqrt{\frac{{{t}^{2}}}{{{t}^{2}}+1}} \end{align}$
Jawaban: E

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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) $6\sqrt{2}$

Penyelesaian: Lihat/Tutup

 
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

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Soal No. 17

Perhatikan gambar berikut!

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) $\frac{c}{a}$

Penyelesaian: Lihat/Tutup

$\sin \alpha =\frac{\text{sisi}\,\text{depan}}{\text{hipotenusa}}=\frac{a}{c}$
Jawaban: B

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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) $\frac{{{q}^{2}}-{{p}^{2}}}{{{q}^{2}}}$

Penyelesaian: Lihat/Tutup

$\sin A=\frac{p}{q}=\frac{de}{mi}$
$sa=\sqrt{m{{i}^{2}}-d{{e}^{2}}}=\sqrt{{{q}^{2}}-{{p}^{2}}}$
$\begin{align} {{\cos }^{2}}A &={{\left( \frac{sa}{mi} \right)}^{2}} \\ &={{\left( \frac{\sqrt{{{q}^{2}}-{{p}^{2}}}}{q} \right)}^{2}} \\ {{\cos }^{2}}A &=\frac{{{q}^{2}}-{{p}^{2}}}{{{q}^{2}}} \end{align}$
Jawaban: E

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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) $-\frac{3}{5}$

Penyelesaian: Lihat/Tutup

$\tan \alpha =\frac{1}{3}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{1}^{2}}+{{3}^{2}}} \\ mi &=\sqrt{10} \end{align}$
$\begin{align} 2\sin \alpha .\cos \alpha &=2.\frac{de}{mi}.\frac{sa}{mi} \\ &=2.\frac{1}{\sqrt{10}}.\frac{3}{\sqrt{10}} \\ &=\frac{6}{10} \\ 2\sin \alpha .\cos \alpha &=\frac{3}{5} \end{align}$
Jawaban: A

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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) $\frac{2p}{\sqrt{1+{{p}^{2}}}}$

Penyelesaian: Lihat/Tutup

$\sin 25{}^\circ =p\Leftrightarrow \frac{de}{mi}=\frac{p}{1}$
$sa=\sqrt{m{{i}^{2}}-d{{e}^{2}}}=\sqrt{1-{{p}^{2}}}$
$\tan 25{}^\circ =\frac{de}{sa}=\frac{p}{\sqrt{1-{{p}^{2}}}}$
Jawaban: C

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