Bank Soal Sudut-Sudut Berelasi dan Pembahasan

Berikut ini adalah Soal-Soal Sudut Berelasi dan Pembahasannya, Sudut-sudut Berelasi adalah sub topik dari materi TRIGONOMETRI pada mata pelajaran Matematika Wajib Kelas 10 Kurikulum 2013. Silahkan dimanfaatkan sebaik mungkin.

Tata Cara Belajar:
Cobalah mengerjakan soal-soal yang tersedia secara mandiri. Setelah itu cocokkanlah jawaban kamu dengan pembahasan yang telah disediakan, dengan cara klik "LIHAT PEMBAHASAN:".

Soal No. 1

Nilai $\sin 50{}^\circ $ = …
(A) $\sin 310{}^\circ $
(B) $\sin 230{}^\circ $
(C) $\cos 110{}^\circ $
(D) $\cos 320{}^\circ $
(E) $\cos 210{}^\circ $

Kita bahas per opsi aja ya. Biar makin mantap:
Opsi A:
$\sin 310{}^\circ =\sin (360{}^\circ -50{}^\circ )=-\sin 50{}^\circ $
Opsi B:
$\sin 230{}^\circ =\sin (180{}^\circ +50{}^\circ )=-\sin 50{}^\circ $
Opsi C:
$\cos 110{}^\circ =\cos (90{}^\circ +20{}^\circ )=-\sin 20{}^\circ $
Opsi D:
$\cos 320{}^\circ =\cos (270{}^\circ +50{}^\circ )=\sin 50{}^\circ $
Opsi E:
$\cos 210{}^\circ =\cos (270{}^\circ -60{}^\circ )=-\sin 60{}^\circ $
Jadi, opsi yang sesuai adalah opsi D.
Jawaban: D

Soal No. 2

Jika diketahui $\cos x=\frac{\sqrt{5}}{5}$ dengan $x$ sudut lancip, maka nilai dari $\cot \left( \frac{\pi }{2}+x \right)$ = …
(A) $-3$
(B) $-2$
(C) $-4$
(D) $-5$
(E) $-6$

$\cos x=\frac{\sqrt{5}}{5}=\frac{sa}{mi}$
$\begin{align} de &=\sqrt{m{{i}^{2}}-s{{a}^{2}}} \\ &=\sqrt{{{5}^{2}}-{{(\sqrt{5})}^{2}}} \\ &=\sqrt{25-5} \\ &=\sqrt{20} \\ de &=2\sqrt{5} \end{align}$
$\begin{align} \cot \left( \frac{\pi }{2}+x \right) &=-\tan x \\ &=-\frac{de}{sa} \\ &=-\frac{2\sqrt{5}}{\sqrt{5}} \\ \cot \left( \frac{\pi }{2}+x \right) &=-2 \end{align}$
Jawaban: B

Soal No. 3

Jika $\sin x=\frac{1}{5}\sqrt{5}$, maka $\cos x-5\cos \left( \frac{\pi }{2}+x \right)+2\sin (\pi -x)$ = …
(A) $-\frac{1}{5}-\frac{1}{5}\sqrt{5}$
(B) $-\sqrt{5}$
(C) $\frac{1}{5}\sqrt{5}$
(D) $\frac{3}{5}\sqrt{5}$
(E) $\frac{9}{5}\sqrt{5}$

$\begin{align} \sin x &=\frac{1}{5}\sqrt{5} \\ \frac{de}{mi} &=\frac{\sqrt{5}}{5} \end{align}$
$\begin{align} sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ &=\sqrt{{{5}^{2}}-{{(\sqrt{5})}^{2}}} \\ &=\sqrt{25-5} \\ &=\sqrt{20} \\ sa &=2\sqrt{5} \end{align}$
$\cos x=\frac{sa}{mi}=\frac{2\sqrt{5}}{5}$
$\cos x-5\cos \left( \frac{\pi }{2}+x \right)+2\sin (\pi -x)$
= $\cos x-5\left( -\sin x \right)+2\sin x$
= $\frac{2\sqrt{5}}{5}+5.\frac{\sqrt{5}}{5}+2.\frac{\sqrt{5}}{5}$
= $\frac{9\sqrt{5}}{5}$
Jawaban: E

Soal No. 4

Jika $\alpha $ di kuadran II dan $\tan \alpha =-\frac{2}{3}$, nilai dari $\frac{\sin (90{}^\circ -\alpha )-\cos (180{}^\circ -\alpha )}{\tan (270{}^\circ +\alpha )+\cot (360{}^\circ -\alpha )}$ = …
(A) $-\frac{2}{13}\sqrt{13}$
(B) $\frac{2}{13}\sqrt{13}$
(C) $\sqrt{13}$
(D) $2\sqrt{13}$
(E) $3\sqrt{13}$

$\tan \alpha =-\frac{2}{3}=\frac{de}{sa}$
$\begin{align} mi &=\sqrt{d{{e}^{2}}+s{{a}^{2}}} \\ &=\sqrt{{{2}^{2}}+{{3}^{2}}} \\ mi &=\sqrt{13} \end{align}$
$\alpha $ di kuadran II maka $\sin \alpha $ bernilai positif.
$\sin \alpha =\frac{de}{mi}=\frac{2}{\sqrt{13}}$
$\frac{\sin (90{}^\circ -\alpha )-\cos (180{}^\circ -\alpha )}{\tan (270{}^\circ +\alpha )+\cot (360{}^\circ -\alpha )}$
= $\frac{\cos \alpha +\cos \alpha }{-\cot \alpha -\cot \alpha }$
= $\frac{2\cos \alpha }{-2\cot \alpha }$
= $\frac{-\cos \alpha }{\cos \alpha /\sin \alpha }$
= $-\sin \alpha $
= $-\frac{2}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}$
= $-\frac{2}{13}\sqrt{13}$
Jawaban: A

Soal No. 5

$\sec 225{}^\circ $ = ….
(A) $-\sqrt{3}$
(B) $-\sqrt{2}$
(C) $-1$
(D) $-\frac{1}{2}\sqrt{3}$
(E) $-\frac{1}{2}\sqrt{2}$

$\begin{align} \sec 225{}^\circ &=\sec (180{}^\circ +45{}^\circ ) \\ &=-\sec 45{}^\circ \\ &=-\frac{1}{\cos 45{}^\circ } \\ &=-\frac{1}{\frac{1}{\sqrt{2}}} \\ \sec 225{}^\circ &=-\sqrt{2} \end{align}$
Jawaban: B

Soal No. 6

${{\cos }^{2}}30{}^\circ -{{\sin }^{2}}135{}^\circ +8\sin 45{}^\circ .\cos 135{}^\circ $ = ….
(A) $-4\frac{1}{4}$
(B) $-3\frac{3}{4}$
(C) $4\frac{1}{4}$
(D) 4
(E) $3\frac{3}{4}$

${{\cos }^{2}}30{}^\circ -{{\sin }^{2}}135{}^\circ +8\sin 45{}^\circ .\cos 135{}^\circ $
= ${{\cos }^{2}}30{}^\circ$ -${{\sin }^{2}}(180{}^\circ -45{}^\circ )$+$8\sin 45{}^\circ \cos (180{}^\circ -45{}^\circ )$
= ${{\cos }^{2}}30{}^\circ$ -${{\sin }^{2}}45{}^\circ$ +$8\sin 45{}^\circ .(-\cos 45{}^\circ )$
= ${{\left( \frac{1}{2}\sqrt{3} \right)}^{2}}-{{\left( \frac{1}{2}\sqrt{2} \right)}^{2}}+8.\frac{1}{2}\sqrt{2}\left( -\frac{1}{2}\sqrt{2} \right)$
= $\frac{3}{4}-\frac{2}{4}-\frac{16}{4}$
= $-\frac{15}{4}$
= $-3\frac{3}{4}$
Jawaban: B

Soal No. 7

Nilai dari: $5\sec 540{}^\circ -4\cos 630{}^\circ +\sin 360{}^\circ +3\cot 450{}^\circ $ adalah …
(A) $-5$
(B) 1
(C) 1,5
(D) 4
(E) 9

$5\sec 540{}^\circ$ -$4\cos 630{}^\circ$ +$\sin 360{}^\circ$ +$3\cot 450{}^\circ $
= $5\sec (360{}^\circ$ +$180{}^\circ )$-$4\cos (360{}^\circ +270{}^\circ )$+$\sin 360{}^\circ$ +$3\cot (360{}^\circ +90{}^\circ )$
= $5\sec 180{}^\circ$ -$4\cos 270{}^\circ$ +$\sin 360{}^\circ$ +$3\cot 90{}^\circ $
= $5\sec 180{}^\circ$ -$4\cos 270{}^\circ$ +$\sin 360{}^\circ$ +$3\cot 90{}^\circ $
= $5\sec (180{}^\circ +0{}^\circ )$-$4\cos (270{}^\circ +0{}^\circ )$+$\sin (360{}^\circ +0{}^\circ )$+$3\cot (90{}^\circ +0{}^\circ )$
= $-5\sec 0{}^\circ$ +$4\sin 0{}^\circ$ +$\sin 0{}^\circ$ -$3\tan 0{}^\circ $
= $-5.1+4.0+0-3.0$
= $-5$
Jawaban: A

Soal No. 8

Jika $\sin 5{}^\circ =p$ maka $\cos 265{}^\circ $ adalah …
(A) $2p$
(B) $p_{{}}^{{}}$
(C) $\frac{1}{2}p$
(D) $-p$
(E) $-2p$

$\begin{align} \cos 265{}^\circ &=\cos (270{}^\circ -5{}^\circ ) \\ &=-\sin 5{}^\circ \\ \cos 265{}^\circ &=-p \end{align}$
Jawaban: D

Soal No. 9

Nilai dari $\frac{\sin 45{}^\circ .\sin 25{}^\circ }{\sin 30{}^\circ .\cos 65{}^\circ }$ = ….
(A) 1
(B) $\frac{1}{2}\sqrt{2}$
(C) $\sqrt{2}$
(D) $\frac{1}{3}\sqrt{6}$
(E) $\sqrt{6}$

$\begin{align} \frac{\sin 45{}^\circ .\sin 25{}^\circ }{\sin 30{}^\circ .\cos 65{}^\circ } &=\frac{\sin 45{}^\circ .\sin 25{}^\circ }{\sin 30{}^\circ .\cos (90{}^\circ -25{}^\circ )} \\ &=\frac{\sin 45{}^\circ .\sin 25{}^\circ }{\sin 30{}^\circ .\sin 25{}^\circ } \\ &=\frac{\sin 45{}^\circ }{\sin 30{}^\circ } \\ &=\frac{\frac{1}{2}\sqrt{2}}{\frac{1}{2}} \\ \frac{\sin 45{}^\circ .\sin 25{}^\circ }{\sin 30{}^\circ .\cos 65{}^\circ } &=\sqrt{2}
\end{align}$
Jawaban: C

Soal No. 10

Dalam segitiga ABC sembarang, nilai $\sin \frac{1}{2}(A+B)$ = ….
(A) $\sin \frac{1}{2}C$
(B) $\sin C$
(C) $\cos \frac{1}{2}C$
(D) $\cos C$
(E) $\cos 2C$

$\begin{align} A+B+C &=180{}^\circ \\ A+B &=180{}^\circ -C \\ \frac{1}{2}(A+B) &=\frac{1}{2}(180{}^\circ -C) \\ \frac{1}{2}(A+B) &=90{}^\circ -\frac{1}{2}C \\ \sin \frac{1}{2}(A+B) &=\sin \left( 90{}^\circ -\frac{1}{2}C \right) \\ \sin \frac{1}{2}(A+B) &=\cos \frac{1}{2}C \end{align}$
Jawaban: C

Soal No. 11

Jika $\sin 35{}^\circ =p$, maka $\cos 55{}^\circ $ = ….
(A) $p$
(B) $1-p$
(C) ${{p}^{2}}$
(D) $1-{{p}^{-1}}$
(E) $\frac{1}{p}$

$\begin{align} \cos 55{}^\circ &=\cos (90{}^\circ -35{}^\circ ) \\ &=\sin 35{}^\circ \\ \cos 55{}^\circ &=p \end{align}$
Jawaban: A

Soal No. 12

Jika $\cos 10{}^\circ =m$ maka $\sin 100{}^\circ $ = ….
(A) $m$
(B) $-m$
(C) ${{m}^{2}}$
(D) $-{{m}^{2}}$
(E) $\sqrt{m}$

$\begin{align} \sin 100{}^\circ &=\sin (90{}^\circ +10{}^\circ ) \\ &=\cos 10{}^\circ \\ \sin 100{}^\circ &=m \end{align}$
Jawaban: A

Soal No. 13

Jika $\sin 35{}^\circ =k$ maka $\cos 125{}^\circ $ = ….
(A) $k$
(B) $-k$
(C) ${{k}^{2}}$
(D) $-{{k}^{2}}$
(E) $\frac{1}{2}p$

$\begin{align} \cos 125{}^\circ &=\cos (90+35) \\ &=-\sin 35 \\ \cos 125{}^\circ &=-k \end{align}$
Jawaban: B

Soal No. 14

$\sin (90{}^\circ +A)+\cos (180{}^\circ -A)+\tan (90{}^\circ +A)$ = …
(A) $2\cos A-\cot A$
(B) $\cot A-2\cos A$
(C) $2\sin A+\cot A$
(D) $\cot A$
(E) $-\cot A$

$\sin (90{}^\circ +A)+\cos (180{}^\circ -A)+\tan (90{}^\circ +A)$
= $\cos A+(-\cos A)+(-\cot A)$
= $-\cot A$
Jawaban: E

Soal No. 15

$\frac{-\sin 45{}^\circ .\sin 15{}^\circ }{\cos 135{}^\circ .\cos 105{}^\circ }$ = ….
(A) $-2$
(B) $-1$
(C) 0
(D) 1
(E) 2

$\begin{align} \frac{-\sin 45{}^\circ .\sin 15{}^\circ }{\cos 135{}^\circ .\cos 105{}^\circ } &=\frac{-\sin 45{}^\circ .\sin 15{}^\circ }{\cos (180{}^\circ -45{}^\circ ).\cos (90{}^\circ +15{}^\circ )} \\ &=\frac{-\sin 45{}^\circ .\sin 15{}^\circ }{-\cos 45{}^\circ .(-\sin 15{}^\circ )} \\ &=\frac{-\sin 45{}^\circ .\sin 15{}^\circ }{-\cos 45{}^\circ .(-\sin 15{}^\circ )} \\ &=-\tan 45{}^\circ \\ \frac{-\sin 45{}^\circ .\sin 15{}^\circ }{\cos 135{}^\circ .\cos 105{}^\circ } &=-1 \end{align}$
Jawaban: B

Soal No. 16

$\left( \sin 20{}^\circ -\cos 110{}^\circ \right)\left( \sin 20{}^\circ +\cos 110{}^\circ \right)$ = ….
(A) $-2$
(B) $-1$
(C) 0
(D) 1
(E) 2

$\cos 110{}^\circ =\cos (90{}^\circ +20{}^\circ )=-\sin 20{}^\circ $
$\left( \sin 20{}^\circ -\cos 110{}^\circ \right)\left( \sin 20{}^\circ +\cos 110{}^\circ \right)$
= $\left( \sin 20{}^\circ +\sin 20{}^\circ \right)\left( \sin 20{}^\circ -\sin 20{}^\circ \right)$
= $(2\sin 20{}^\circ ).0$
= 0
Jawaban: C

Soal No. 17

Jika $\sin 48,59{}^\circ =0,75$ maka $\cos 138,59{}^\circ $ = …
(A) 0,75
(B) 0,25
(C) 0,15
(D) $-075$
(E) $-0,25$

$\begin{align} \cos 138,59{}^\circ &=\cos (90{}^\circ +48,59{}^\circ ) \\ &=-\sin 48,59{}^\circ \\ \cos 138,59{}^\circ &=-0,75 \end{align}$
Jawaban: D

Soal No. 18

$\sin 230{}^\circ $ = ….
(A) $\sin 50{}^\circ $
(B) $\cos 50{}^\circ $
(C) $-\sin 50{}^\circ $
(D) $-\cos 50{}^\circ $
(E) $\tan 50{}^\circ $

$\sin 230{}^\circ =\sin (180{}^\circ +50{}^\circ )=-\sin 50{}^\circ $
Jawaban: C

Soal No. 19

$\cos 205{}^\circ $ = ….
(A) $\sin 25{}^\circ $
(B) $\cos 25{}^\circ $
(C) $-\sin 25{}^\circ $
(D) $-\cos 25{}^\circ $
(E) $\tan 25{}^\circ $

$\cos 205{}^\circ =\cos (180{}^\circ +25{}^\circ )=-\cos 25{}^\circ $
Jawaban: D

Soal No. 20

$\sin 255{}^\circ $ = ….
(A) $\sin 75{}^\circ $
(B) $\sin 65{}^\circ $
(C) $-\sin 75{}^\circ $
(D) $-\sin 65{}^\circ $
(E) $\sin 55{}^\circ $

$\sin 255{}^\circ =\sin (180{}^\circ +75{}^\circ )=-\sin 75{}^\circ $
Jawaban: C

Soal No. 21

$\cos 265{}^\circ $ = …
(A) $\cos 85{}^\circ $
(B) $\cos 75{}^\circ $
(C) $\cos 65{}^\circ $
(D) $-\cos 85{}^\circ $
(E) $-\cos 75{}^\circ $

$\cos 265{}^\circ =\cos (180{}^\circ +85{}^\circ )=-\cos 85{}^\circ $
Jawaban: D

Soal No. 22

$\tan 190{}^\circ $ = ….
(A) $\tan 10{}^\circ $
(B) $\cot 10{}^\circ $
(C) $-\tan 10{}^\circ $
(D) $-\cot 10{}^\circ $
(E) $\tan 20{}^\circ $

$\tan 190{}^\circ =\tan (180{}^\circ +10{}^\circ )=\tan 10{}^\circ $
Jawaban: A

Soal No. 23

Jika $\cos 20{}^\circ =m$ maka $\sin 250{}^\circ $ = ….
(A) $m$
(B) $-m$
(C) ${{m}^{2}}$
(D) $-{{m}^{2}}$
(E) $\frac{1}{m}$

$\begin{align} \sin 250{}^\circ &=\sin (270{}^\circ -20{}^\circ ) \\ &=-\cos 20{}^\circ \\ \sin 250{}^\circ &=-m \end{align}$
Jawaban: B

Soal No. 24

Jika $\cos 256{}^\circ =k$ maka $\sin 14{}^\circ $ = ….
(A) $k$
(B) $-k$
(C) ${{k}^{2}}$
(D) $-{{k}^{2}}$
(E) $\sqrt{k}$

$\begin{align} \cos 256{}^\circ &=k \\ \cos (270{}^\circ -14{}^\circ ) &=k \\ -\sin 14{}^\circ &=k \\ \sin 14{}^\circ &=-k \end{align}$
Jawaban: B

Soal No. 25

Jika $\tan 27{}^\circ =0,51$ maka $\cot 243{}^\circ $ = ….
(A) 0,61
(B) 0,51
(C) 0,49
(D) $-0,49$
(E) $-0,51$

$\begin{align} \cot 243{}^\circ &=\cot (270{}^\circ -27{}^\circ ) \\ &=\tan 27{}^\circ \\ \cot 243{}^\circ &=0,51 \end{align}$
Jawaban: B

Soal No. 26

$\sin 210{}^\circ $ = …
(A) $\frac{1}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}\sqrt{3}$
(D) $-\frac{1}{2}\sqrt{3}$
(E) 1

$\begin{align} \sin 210{}^\circ &=\sin (180{}^\circ +30{}^\circ ) \\ &=-\sin 30{}^\circ \\ \sin 210{}^\circ &=-\frac{1}{2} \end{align}$
Jawaban: B

Soal No. 27

$\cos 240{}^\circ $ = ….
(A) $\frac{1}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}\sqrt{3}$
(D) $-\frac{1}{2}\sqrt{3}$
(E) $-1$

$\begin{align} \cos 240{}^\circ &=\cos (180{}^\circ +60{}^\circ ) \\ &=-\cos 60{}^\circ \\ \cos 240{}^\circ &=-\frac{1}{2} \end{align}$
Jawaban: B

Soal No. 28

$\tan 225{}^\circ $ = ….
(A) 1
(B) $-1$
(C) $\sqrt{3}$
(D) $-\sqrt{3}$
(E) $\frac{1}{2}\sqrt{2}$

$\begin{align} \tan 225{}^\circ &=\tan (180{}^\circ +45{}^\circ ) \\ &=\tan 45{}^\circ \\ \tan 225{}^\circ &=1 \end{align}$
Jawaban: A

Soal No. 29

$\sin (180{}^\circ +x)-\cot (90{}^\circ +x)-\tan (180{}^\circ +x)$ = ….
(A) $\sin x$
(B) $-\sin x$
(C) $\sin x-2\tan x$
(D) $-\sin x-2\tan x$
(E) 0

$\sin (180{}^\circ +x)-\cot (90{}^\circ +x)-\tan (180{}^\circ +x)$
= $-\sin x-(-\tan x)-\tan x$
= $-\sin x$
Jawaban: B

Soal No. 30

$\sin (270{}^\circ -A)-\cos (180{}^\circ -A)+\cot (270{}^\circ -A)$ = …
(A) $-2\cos A-\tan A$
(B) $-\cos A+\tan A$
(C) $2\cos A+\tan A$
(D) $\tan A$
(E) $-\tan A$

$\sin (270{}^\circ -A)-\cos (180{}^\circ -A)+\cot (270{}^\circ -A)$
= $-\cos A-(-\cos A)+\tan A$
= $\tan A$
Jawaban: D

Soal No. 31

$\sin x=0,1$ maka $\sin \left( x+\frac{1}{2}\pi \right)+\cos (\pi +x)$ = ….
(A) $-0,2$
(B) $-0,1$
(C) 0
(D) 0,2
(E) $-\frac{3}{5}\sqrt{11}$

$\begin{align} \sin x &=0,1 \\ \frac{de}{mi} &=\frac{1}{10} \end{align}$
$\begin{align} sa &=\sqrt{m{{i}^{2}}-d{{e}^{2}}} \\ &=\sqrt{{{10}^{2}}-{{1}^{2}}} \\ &=\sqrt{99} \\ sa &=3\sqrt{11} \end{align}$
$\begin{align} \sin \left( x+\frac{1}{2}\pi \right)+\cos (\pi +x) &=-\cos x-\cos x \\ &=-2\cos x \\ &=-2.\frac{sa}{mi} \\ &=-2.\frac{3\sqrt{11}}{10} \\ \sin \left( x+\frac{1}{2}\pi \right)+\cos (\pi +x) &=-\frac{3\sqrt{11}}{5} \end{align}$
Jawaban: E

Soal No. 32

Jika $\tan 24{}^\circ =k$ maka $\cos 294{}^\circ $ = ….
(A) $\frac{k}{\sqrt{1-{{k}^{2}}}}$
(B) $\frac{k}{\sqrt{{{k}^{2}}+1}}$
(C) $\sqrt{1+{{k}^{2}}}$
(D) $\sqrt{1-{{k}^{2}}}$
(E) 1

$\begin{align} \tan 24{}^\circ &=k \\ \frac{de}{sa} &=\frac{k}{1} \end{align}$
$mi=\sqrt{d{{e}^{2}}+s{{a}^{2}}}=\sqrt{{{k}^{2}}+1}$
$\begin{align} \cos 294{}^\circ &=\cos (270{}^\circ +24{}^\circ ) \\ &=\sin 24{}^\circ \\ &=\frac{de}{mi} \\ \cos 294{}^\circ &=\frac{k}{\sqrt{{{k}^{2}}+1}} \end{align}$
Jawaban: B

Soal No. 33

$\cos (270{}^\circ +A)+\cos (270{}^\circ -A)+\tan (270{}^\circ +A)$ = …
(A) $-2\sin A-\cot A$
(B) $-2\sin A+\cot A$
(C) $2\sin A+\cot A$
(D) $\cot A$
(E) $-\cot A$

$\cos (270{}^\circ +A)+\cos (270{}^\circ -A)+\tan (270{}^\circ +A)$
= $\sin A+(-\sin A)+(-\cot A)$
= $-\cot A$
Jawaban: E

Soal No. 34

$\sin 280{}^\circ $ = ….
(A) $\sin 80{}^\circ $
(B) $-\sin 80{}^\circ $
(C) $\cos 80{}^\circ $
(D) $-\cos 80{}^\circ $
(E) $-\cos 280{}^\circ $

$\sin 280{}^\circ =\sin (360{}^\circ -80{}^\circ )=-\sin 80{}^\circ $
Jawaban: B

Soal No. 35

Jika $\sin (360{}^\circ -A)=p$ maka $\sin A$ = ….
(A) $1-p$
(B) $-p$
(C) $p$
(D) $1+p$
(E) $1+{{p}^{2}}$

$\begin{align} \sin (360{}^\circ -A) &=p \\ -\sin A &=p \\ \sin A &=-p \end{align}$
Jawaban: B

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