Soal Perbandingan Trigonometri 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/Tutup".

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 $

Penyelesaian: Lihat/Tutup

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

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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) $-6$

Penyelesaian: Lihat/Tutup

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

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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) $\frac{9}{5}\sqrt{5}$

Penyelesaian: Lihat/Tutup

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

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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) $3\sqrt{13}$

Penyelesaian: Lihat/Tutup

$\tan \alpha =-\frac{2}{3}=\frac{de}{sa}$
$\begin{align} mi &= \sqrt{de^2+sa^2} \\ &= \sqrt{{{2}^{2}}+{{3}^{2}}} \\ mi &= \sqrt{13} \end{align}$
$\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

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Soal No. 5

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

Penyelesaian: Lihat/Tutup

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

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

Penyelesaian: Lihat/Tutup

${{\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

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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

Penyelesaian: Lihat/Tutup

$5\sec 540^\circ$ -$4\cos 630^\circ$ +$\sin 360^\circ$ +$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

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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) $-2p$

Penyelesaian: Lihat/Tutup

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

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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) $\sqrt{6}$

Penyelesaian: Lihat/Tutup

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

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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) $\cos 2C$

Penyelesaian: Lihat/Tutup

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

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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) $\frac{1}{p}$

Penyelesaian: Lihat/Tutup

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

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Soal No. 12

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

Penyelesaian: Lihat/Tutup

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

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Soal No. 13

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

Penyelesaian: Lihat/Tutup

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

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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) $-\cot A$

Penyelesaian: Lihat/Tutup

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

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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

Penyelesaian: Lihat/Tutup

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

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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

Penyelesaian: Lihat/Tutup

$\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

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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) $-0,25$

Penyelesaian: Lihat/Tutup

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

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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 $

Penyelesaian: Lihat/Tutup

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

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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 $

Penyelesaian: Lihat/Tutup

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

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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 $

Penyelesaian: Lihat/Tutup

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

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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 $

Penyelesaian: Lihat/Tutup

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

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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 $

Penyelesaian: Lihat/Tutup

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

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Soal No. 23

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

Penyelesaian: Lihat/Tutup

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

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Soal No. 24

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

Penyelesaian: Lihat/Tutup

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

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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) $-0,51$

Penyelesaian: Lihat/Tutup

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

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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

Penyelesaian: Lihat/Tutup

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

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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) $-1$

Penyelesaian: Lihat/Tutup

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

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Soal No. 28

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

Penyelesaian: Lihat/Tutup

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

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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

Penyelesaian: Lihat/Tutup

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

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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) $-\tan A$

Penyelesaian: Lihat/Tutup

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

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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) $-\frac{3}{5}\sqrt{11}$

Penyelesaian: Lihat/Tutup

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

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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

Penyelesaian: Lihat/Tutup

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

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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) $-\cot A$

Penyelesaian: Lihat/Tutup

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

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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 $

Penyelesaian: Lihat/Tutup

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

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Soal No. 35

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

Penyelesaian: Lihat/Tutup

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

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