If $$\sqrt{\sin x}+\cos x=0$$, then $$\sin x=$$

$$\displaystyle \frac{\sqrt{5}-1}{2}$$

$$\displaystyle \frac{\sqrt{5}+1}{2}$$

$$\displaystyle \frac{\sqrt{5}+1}{8}$$

$$\displaystyle \frac{\sqrt{5}-1}{8}$$

MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 3

$\tan \theta +\cot \theta =3$ , then

$\tan^{4}\theta+\cot^{4}\theta=$

0%
47
0%
162
0%
24
0%
48

Explanation

Given

$\tan \theta +\cot \theta =3$

$\Rightarrow (\tan \theta +\cot \theta )^{2} = \tan ^{2}\theta +\cot^{2}\theta +2=9$

$\Rightarrow \tan ^{2}\theta +\cot ^{2}\theta =7$

$\Rightarrow \tan ^{4}\theta +\cot ^{4} \theta +2=49$

$\Rightarrow \tan ^{4}\theta +\cot ^{4}\theta =47$

$\cos x + \sec x = - 2$ for a positive odd integer

$n$ then

$\cos^nx + \sec^nx$ is

0%
$1$
0%
$-1$
0%
$-2$
0%
$2$

Explanation

$\cos x + \sec x = - 2$

$\cos x + \displaystyle\frac{1}{\cos x} =-2$

$\cos ^2x +1 =-2 \cos x$

$\cos ^2x +1 +2 \cos x =0$

This is of the form

$a^2+b^2+2ab=(a+b)^2$

$\Rightarrow (\cos x +1)^2 = 0$

$\Rightarrow (\cos x +1) = 0$

$\Rightarrow \cos x =-1$

$\implies \sec x=-1$

$n$ is odd,

$\cos ^nx + \sec ^nx =-1+(-1)= - 2$

$n$ is even,

$\cos ^nx + \sec ^nx = 1+1=2$

$\sqrt{\sin x}+\cos x=0$ , then

$\sin x=$

0%
$\displaystyle \frac{\sqrt{5}+1}{2}$
0%
$\displaystyle \frac{\sqrt{5}+1}{8}$
0%
$\displaystyle \frac{\sqrt{5}-1}{8}$
0%
$\displaystyle \frac{\sqrt{5}-1}{2}$

Explanation

${\textbf{Step -1: Rewrite the given equation as quadratic equation in Sinx by applying Standard Trigonometric Identity}}$

${\text{Consider,}}$

$\sqrt {\sin x} + \cos x = 0$

$\Rightarrow \sqrt {\sin x} = - \cos x$

${\text{Squaring on both sides, we get}}$

${\left( {\sqrt {\sin x} } \right)^2} = {\left( { - \cos x} \right)^2}$

$\Rightarrow \sin x = {\cos ^2}x \ldots \left( 1 \right)$

${\text{As we know that,}}$

${\sin ^2}x + {\cos ^2}x = 1$

$\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$ ${\textbf{(Substitute this value in equation }}\left( 1 \right))$

$\Rightarrow \sin x = 1 - {\sin ^2}x$

$\Rightarrow {\sin ^2}x + \sin x - 1 = 0 \ldots \left( 2 \right)$

${\textbf{Step -2: Solving equation}}\left( \mathbf 2 \right)$ ${\textbf{for}}$ $\textbf {sin} \mathbf{x.}$

${\text{If }}$ $a{x^2} + bx + c = 0$ ${\text{is a quadratic equation then roots of the equation are,}}$

$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

${\text{So, comparing it with equation }}\left( 2 \right)$ ${\text{we get,}}$

$a = 1,b = 1,c = - 1$

$\therefore \sin x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$

$\Rightarrow \sin x = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$

$\Rightarrow \sin x = \dfrac{{ - 1 - \sqrt 5 }}{2}$ ${\text{or}}$ $\sin x = \dfrac{{ - 1 + \sqrt 5 }}{2}$

$\Rightarrow \sin x = \dfrac{{\sqrt 5 - 1}}{2}$

$\because \dfrac{{ - 1 - \sqrt 5 }}{2} < - 1,{\text{Therefore, the other value is unacceptable}}{\text{.}}$

${\textbf{Hence, option (D)}}$ $\mathbf{\dfrac{{\sqrt 5 - 1}}{2}}$ ${\textbf{is the correct answer.}}$

Eiminate

$\theta$ from

$x= 1+\tan\theta,y= 2+\cot\theta$

0%
$xy+1=x+y$
0%
$xy+2=2x+y$
0%
$xy+1=2x+y$
0%
$1=2y+x$

Explanation

We know,

$\tan \theta \times \cot \theta = 1$
Substituting the corresponding values,

$(x - 1)(y - 2) = 1$

$xy - 2x - y + 2 = 1$
Or,

$xy + 1 = 2x + y$
Hence, option 'C' is correct.

The maximum value of

$\cos x\,\left(\displaystyle \dfrac{\cos x}{1-\sin x}+\dfrac{1-\sin x}{\cos x}\right)$ is

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$\displaystyle f(x)=\cos x(\frac{\cos ^{2}x+1+\sin ^{2}x-2\sin x}{\cos x(1-\sin x)})$

$\displaystyle =\cos x(\frac{2(1-\sin x)}{\cos x(1-\sin x)}) =\cos x(\frac{2}{\cos x})=2$

The value of the expression

$\text{cosec }(75^0+\theta) - \sec (15^0 - \theta) - \tan (55^0 + \theta) + \cot (35^0 - \theta)$ , is

0%
$-1$
0%
$0$
0%
$1$
0%
$\dfrac{3}{2}$

Explanation

Given:

$E = \text{cosec }(75^0 + \theta) - \sec (15^0 - \theta) - \tan (55^0 + \theta) + \cot (35^0 - \theta)$

As, we know that,

$\text{cosec }\theta = \sec(90^0-\theta)$ and

$\tan\theta = \cot(90^0-\theta)$

Using these, we get

$E=\sec(90^0-75^0-\theta)-\sec(15^0-\theta)-\cot(90^0-55^0-\theta)+\cot(35^0-\theta)$

$\Rightarrow E=\sec(15^0-\theta)-\sec(15^0-\theta)-\cot(35^0-\theta)+\cot(35^0-\theta) = 0$

Hence, option B is correct.

$\sin \left ( 45^{0}+\theta \right )-\cos\left ( 45^{0}-\theta \right )$ is equal to

0%
$2 \cos \theta$
0%
$0$
0%
$2 \sin \theta$
0%
$1$

Explanation

$\sin (45^0+\theta)-\cos (45^0-\theta)$

$=sin(45^0+\theta)-sin(90^0-(45^0-\theta))$

$=sin (45^0+\theta)-sin( 45^0+\theta)$

$=0$

$\displaystyle \sec\alpha=5x+\frac{1}{20x}$ , then

$\sec\alpha+\tan\alpha$ is equal to

0%
$5x$
0%
$\displaystyle \frac{1}{20x}$
0%
$10x$ or $-\displaystyle \frac{1}{10x}$
0%
$10\displaystyle x$ or $\dfrac{1}{10x}$

Explanation

Let

$\sec\alpha + \tan\alpha = a$ .....

$(i)$

It is known that,

$\sec^{2}\alpha - \tan^{2}\alpha=1$

$(\sec\alpha+\tan\alpha)(\sec\alpha-\tan\alpha)=1$

So we get

$\sec\alpha - \tan\alpha = \dfrac{1}{a}$ .....

$(ii)$

Adding equations

$(i)$ and

$(ii)$ , we get

$2\sec\alpha = a + \dfrac{1}{a}$

$\Rightarrow \sec\alpha = \dfrac{1}{2}\left(a + \dfrac{1}{a}\right)$

Put the value of

$\sec\alpha$ in the given equation.

$\Rightarrow 5x + \dfrac{1}{20x}=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)$

$\Rightarrow \dfrac{1}{2}\left(10x + \dfrac{1}{10x}\right)=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)$
Solving for

$a$ we get

$a = 10x$ or

$a = \dfrac{1}{10x}$

Hence option (D) is correct

$x.cos A = 1$ and

$tan A = y$ , then

$x^{2} - y^{2}$ is equal to :

0%
$\tan A$
0%
$1$
0%
$sinA$
0%
$-\tan A$

Explanation

$x\cos { A } =1,\tan { A } =y$

$\Rightarrow x=\sec { A }$

Now,

${ x }^{ 2 }-{ y }^{ 2 }=\sec ^{ 2 }{ A } -\tan ^{ 2 }{ A }$

$=1$

Hence, the answer is

$1.$

In the given figure,

$\triangle ABC$ is right angled at B and

$\cot A = \dfrac{3}{4}$ . If

$AC =10$ cm, then the length of

$AB$ is:

0%
$3$ cm
0%
$4$ cm
0%
$6$ cm
0%
$8$ cm

Explanation

$\cot { A } =\dfrac { 3 }{ 4 } =\dfrac { AB }{ BC }$

$\Rightarrow \dfrac { AB }{ BC } =\dfrac { 3 }{ 4 }$

$\Rightarrow AB:BC=3:4$

$\Rightarrow$ Let

$AB=3x\;\&\; BC=4x$

In right angle

$\triangle ABC,$

$\Rightarrow { AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$

$\Rightarrow { \left( 10 \right) }^{ 2 }={ \left( 3x \right) }^{ 2 }+{ \left( 4x \right) }^{ 2 }$

$\Rightarrow 100= \left( 9x^2 \right) + \left( 16x^2 \right)$

$\Rightarrow 25{ x }^{ 2 }=100$

${ \Rightarrow x }^{ 2 }=4$

$\Rightarrow x=2$

$\Rightarrow AB=3x=6cm$

Hence, the answer is

$6cm$

The value of

$sin^{2}30^{\circ}$ -

$cos^{2}30^{\circ}$ is:

0%
$-\dfrac{1}{2}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$

Explanation

$\sin ^{ 2 }{ 30° } -\cos ^{ 2 }{ 30° }$

$= { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }$

$=\dfrac { 1 }{ 4 } -\dfrac { 3 }{ 4 }$

$=-\dfrac { 2 }{ 4 }$

$=-\dfrac { 1 }{ 2 }$

Hence, the answer is

$-\dfrac { 1 }{ 2 }.$

$\cos\theta =\dfrac{2}{3}$ , then the value of

$2\sec\theta +4\tan^{2}\theta -7$ is equal to

0%
$1$
0%
$0$
0%
$3$
0%
$4$

Explanation

$\cos { \theta } =\dfrac { 2 }{ 3 } \Rightarrow \sec { \theta } =\dfrac { 1 }{ \cos { \theta } } =\dfrac { 3 }{ 2 }$

$\Rightarrow 2\sec { \theta } +4\tan ^{ 2 }{ \theta } -7=2\times \dfrac { 3 }{ 2 } +4\left( \sec ^{ 2 }{ \theta } -1 \right) -7$

$=3+4\left( \dfrac { 9 }{ 4 } -1 \right) -7$

$=4\times \dfrac { 5 }{ 4 } -4$

$=5-4$

$=1$

Hence, the answer is

$1.$

$3\cos\theta =2\sin\theta$ , then the value of

$\dfrac{4\sin\theta -3\cos\theta }{2\sin\theta +6\cos\theta }$ is:

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$

Explanation

Given that:

$3\cos { \theta } =2\sin { \theta }$

$\dfrac { \cos { \theta } }{ \sin { \theta } } =\dfrac { 2 }{ 3 }$

$\cot { \theta } =\dfrac { 2 }{ 3 }$

Now,

$\dfrac { 4\sin { \theta } -3\cos { \theta } }{ 2\sin { \theta } +6\cos { \theta } } =\dfrac { \sin { \theta } \left[ 4-3\dfrac { \cos { \theta } }{ \sin { \theta } } \right] }{ \sin { \theta } \left[ 2+6\dfrac { \cos { \theta } }{ \sin { \theta } } \right] }$

Divided and multiplied the numerator and denominator by

$\sin \theta$ .

$=\dfrac { 4-3\cot { \theta } }{ 2+6\cot { \theta } }$

$=\dfrac { 4-\left( 3\times \dfrac { 2 }{ 3 } \right) }{ 2+\left( 6\times \dfrac { 2 }{ 3 } \right) }$

$=\dfrac { 4-2 }{ 2+4 }$

$=\dfrac { 2 }{ 6 }$

$=\dfrac { 1 }{ 3 }.$

Hence, the answer is

$=\dfrac { 1 }{ 3 } .$

$\tan A =\cfrac{5}{12}$ , find the value of

$(\sin A+ \cos A) \times \sec A$ :

0%
$\dfrac{6}{13}$
0%
$\dfrac{7}{12}$
0%
$\dfrac{17}{12}$
0%
$\dfrac{12}{17}$

Explanation

$\tan { A } =\dfrac { 5 }{ 12 }$

$\left( \sin { A } +\cos { A } \right) \times \sec { A }$

$=\sin { A } .\sec { A } +\cos { A } .\sec { A }$

$=\tan { A } +1$

$=\cfrac { 5 }{ 12 } +1$

$=\cfrac { 17 }{ 12 }$

The value of tan

$1^{\circ}.tan2^{\circ}.tan3^{\circ}.......... tan89^{\circ}$ is :

0%
0
0%
1
0%
2
0%
$\dfrac{1}{2}$

Explanation

$\tan \theta = \cot(90^{\circ}-\theta)$

$\therefore \tan1^{\circ}= \cot89^{\circ}$

$\cot89^{\circ}=\dfrac{1}{\tan89^{\circ} }$

Similarly all the term will cancel out each other, except

$\tan45^{\circ}$ , whose value is equal to 1.

If sin

$\Theta$ = cos

$\Theta$ , then the value of

$\Theta$ is :

0%
$0^{\circ}$
0%
$45^{\circ}$
0%
$60^{\circ}$
0%
$30^{\circ}$

Explanation

For

$\theta =45°,\sin { \theta } \& \cos { \theta }$

$\Rightarrow \sin { \theta } =\cos { \theta }$

$\Rightarrow \tan { \theta } =1$

$\Rightarrow \tan { \theta } =tan 45^0$

$\Rightarrow { \theta } =45^0$

The value of

$\dfrac{tan 45^{\circ}}{sin30^{\circ}+cos 60^{\circ}}$ is :

0%
$\dfrac{1}{\sqrt{2}}$
0%
1
0%
$\dfrac{1}{2}$
0%
$\sqrt{2}$

Explanation

$\dfrac { \tan { 45° } }{ \sin { 30° } +\cos { 60° } }$

$=\dfrac { 1 }{ \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } }$

$=1$

Hence, the answer is

$1.$

Simplify:

$\displaystyle \dfrac {\cos 60^0+\sin 60^0}{\cos 60^0-\sin 60^0}$

0%
$\sqrt 3-2$
0%
$\sqrt 3+2$
0%
$-(\sqrt 3+2)$
0%
$1$
0%
$0$

Explanation

$\displaystyle \dfrac {\cos 60^0+\sin 60^0}{\cos 60^0-\sin 60^0}=\dfrac {\dfrac {1}{2}+\dfrac {\sqrt 3}{2}}{\dfrac {1}{2}-\dfrac {\sqrt 3}{2}}$

$=\dfrac {1+\sqrt 3}{1-\sqrt 3}\times \dfrac {1+\sqrt 3}{1+\sqrt 3}$

$=\dfrac {4+2\sqrt 3}{1-3}=-(2+\sqrt 3)$

In the figure given below,

$\Delta ABC$ is right angled at B and tan

$A=\dfrac{4}{3}.$ If

$AC=15$ cm, then the length of BC is :

0%
$4$ cm
0%
$3$ cm
0%
$12$ cm
0%
$9$ cm

Explanation

$\tan A=\dfrac43$

$\Rightarrow AB=\dfrac34BC$

Now, in

$\triangle ABC$

$AB^2+BC^2=AC^2$

$\Bigg(\dfrac34BC\Bigg)^2+BC^2=15^2$

$\Rightarrow \dfrac{25}{16}BC^2=15^2$

$\Rightarrow BC^2=\dfrac{15^2\times 4^2}{5^2}$

$\Rightarrow BC=12$

Therefore, Answer is

$12\ cm$

Which one of the following is correct ?

0%
$sec^2\alpha=1-tan^2 \alpha$
0%
$sin^2\alpha=1+cos^2\alpha$
0%
$tan \alpha cot \alpha=1$
0%
None of these

Explanation

Only statement

$(C)$ is correct.

$\Rightarrow \tan { \alpha } \cot { \alpha } =1$

$\Rightarrow L.H.S=\tan { \alpha } \cot { \alpha }$

$=\dfrac { 1 }{ \cot { \alpha } } \cot { \alpha }$

$=1=R.H.S.$

Hence, the answer is

$\tan { \alpha } \cot { \alpha }=1.$

$\tan 9^o\times \tan 27^o\times \tan 63^o\times \tan 81^o=$

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$\tan { { 9 }^{ \circ }.\tan { { 27 }^{ \circ }.\tan { { 63 }^{ \circ }.\tan { { 81 }^{ \circ } } } } } \\ =\tan { \left( { 90 }^{ \circ }-{ 81 }^{ \circ } \right) .\tan { \left( { 90 }^{ \circ }-{ 63 }^{ \circ } \right) .\tan { { 63 }^{ \circ }.\tan { { 81 }^{ \circ } } } } } \\ =\cot { { 81 }^{ \circ }.\cot { { 63 }^{ \circ }. } \tan { { 63 }^{ \circ }.\tan { { 81 }^{ \circ } } } } \\ =1$

$\tan30^0=x, \tan 45^0=y, \tan 60^0=z,$ then which of the following is/are correct?

0%
$x+y=z$
0%
$xy=z$
0%
$xz=y$
0%
$y+z=x$

Explanation

$\displaystyle x=\tan { { 30 }^{ \circ }=\frac { 1 }{ \sqrt { 3 } }, \\ y=\tan { { 45 }^{ \circ }=1,\\ z=\tan { { 60 }^{ \circ }=\sqrt { 3 } } } }$

$xz=\dfrac { \sqrt { 3 } }{ \sqrt { 3 } } =1=y$

$x + y = \dfrac1{\sqrt3} + 1 = \dfrac{1+\sqrt3}{\sqrt3} ≠ \sqrt3 = z$

$y+z = 1+\sqrt3 ≠\dfrac1{\sqrt3} = x$

The value of x from the equation

$\dfrac{x}{cos45^0.sin60^0}=tan^2 60^0 - tan^2 30^0$ , is

0%
3
0%
2
0%
1
0%
None of these

Explanation

$\displaystyle \frac{x}{cos45^0\:sin60^0}=tan^260^0-tan^230^0$

$\displaystyle \frac{x}{\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}}=3-\frac{1}{3}$

$\displaystyle \frac{2\sqrt{2}x}{\sqrt{3}}=\frac{8}{3}$

$\displaystyle x=\frac{2\sqrt{2}}{\sqrt{3}}$

Hence the answer is D

The value of

$\sec$

$(90^0 - \theta) \sin \theta$ is

0%
$\sec \theta$
0%
$\csc \theta$
0%
$\tan \theta$
0%
$1$

Explanation

$\sec (90^0-\theta)\sin\theta$

$=\csc\theta\:\sin\theta$

$=\dfrac{1}{\sin\theta}\sin\theta$

$=1$
Hence answer is D

The value of

$\sin^2q. \cos^2q (\sec^2q+\text cosec^2q)$ is

0%
$2$
0%
$4$
0%
$1$
0%
$0$

Explanation

Given:

$\sin^2q \cos^2q (\sec^2q+\text{cosec}^2q)$

$=\sin^2q \cos^2q (\dfrac {1}{\cos^2q}+\cfrac {1}{\sin^2q})$

$=\sin^2q \cos^2q(\dfrac{\sin^2q+\cos^2q}{\sin^2q\cos^2q})$

$=\sin^2q+\cos^2q=1$

The value of

$\sin^6q+\cos^6q+3 \sin^2q \cos^2q$ is-

0%
$0$
0%
$1$
0%
$2\sin q \cos q$
0%
$\sin^4q +\cos^4q$

Explanation

$\sin^6q+\cos^6q+3 \sin^2q \cos^2q$

$\Rightarrow (\sin^2q+\cos^2q)(\sin^4q+\cos^4q-\sin^2q \cos^2q)+3 \sin^2q \cos^2q$

$\Rightarrow \sin^4q+\cos^4q+2\sin^2q \cos^2q$

$\Rightarrow (\sin^2q+\cos^2q)^2=1$ .

$\left [ \cfrac{1+tan \theta}{1+cot \theta} \right ]^{4}$ is equal to

0%
$\left ( \cfrac{1+tan^{2} \theta}{1+cot^{2} \theta} \right )^{2}$
0%
$\left ( \cfrac{1+cot \theta}{1+tan \theta} \right )^{2}$
0%
$\left ( \cfrac{1+tan \theta}{1-cot \theta} \right )^{2}$
0%
$\left ( \cfrac{1-cot^{2} \theta}{1+tan^{2} \theta} \right )^{2}$

Explanation

$\left [ \cfrac{1+tan \theta}{1+cot \theta} \right ]^{4}$

$=\left(\dfrac{tan\theta(1+tan\theta)}{1+tan\theta}\right)\:^4$ ...(

$1/cot\theta=tan\theta$ )

$=(tan\theta)\:^4$

$= \tan ^2 \theta \dfrac{\sec ^2 \theta}{cosec ^2 \theta}$

$=\left(\dfrac{tan^2\theta\:(1+tan^2\theta\:)}{(1+cot^2 \theta\:)}\right)$

$=\left(\dfrac{(1+tan^2\theta\:)}{(1+cot^2\theta\:)}\right)\:^2$

Hence the answer is A

The value of

$32 { \cot }^{ 2 }{ 45 }^{ \circ }-8 { \sec }^{ 2 }{ 60 }^{ \circ }+8 { \cos }^{ 3 }{ 30 }^{ \circ }\\$

0%
$3\sqrt{3}$
0%
$2\sqrt{3}$
0%
$\sqrt{3}$
0%
$6\sqrt{3}$

Explanation

$32 { \cot }^{ 2 }{ 45 }^{ \circ }-8 { \sec }^{ 2 }{ 60 }^{ \circ }+8 { \cos }^{ 3 }{ 30 }^{ \circ }\\ =32\times 1-8\times { 2 }^{ 2 }+8\times \dfrac { 3\sqrt { 3 } }{ 8 } \\ =32-32+3\sqrt { 3 } \\ =3\sqrt { 3 }$

$cot\theta + tan\theta =$

0%
$\dfrac {1}{sin\theta cos\theta}$
0%
$sec\theta cosec\theta$
0%
$cot\theta sec^2\theta$
0%
all of these

Explanation

$\cot\theta +\tan\theta\\$

$=\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}$

$=\dfrac{\cos ^2\theta +\sin^2\theta}{\sin\theta \cos\theta}$

$=\dfrac{1}{\sin\theta \cos\theta}$

$cosec\theta \sec\theta$

$=\dfrac{1}{\sin\theta \cos\theta}$

$\cot\theta \sec^2\theta$

$=\dfrac{\cos \theta}{\sin\theta}\times \dfrac{1}{\cos^2\theta}=\dfrac{1}{\sin\theta \cos\theta}$ .

In the given figure, the side

$PQ$ (in cm) is

0%
$25$
0%
$50$
0%
$75$
0%
$80$

Explanation

In the given triangle

$AB=BP$
Hence,

$\triangle ABP$ is an isosceles triangle
Therefore,

$\angle PAB=\angle APB$
Hence,

$\angle BPQ=90^0-(\angle PAB+\angle APB)$

$=(90^0-15^0-15^0)=60^0$
Consider

$\triangle BPQ$

$\cos60^0=\dfrac{1}{2}=\dfrac{PQ}{BP}$

$=\dfrac{PQ}{100}$
Hence,

$PQ=50$ cm
The answer is option B.

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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