MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 8

The value of

$\displaystyle \frac { 2\cos { { 67 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } } -\sin { { 90 }^{ o } }$ is :

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

Explanation

$\displaystyle \frac { 2\cos { { 67 }^{ o } } }{ \sin { { 23 }^{ o } } } +\frac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } } -\sin { { 90 }^{ o } }$

$\displaystyle =\frac { 2\cos { { 67 }^{ o } } }{ \cos { \left( { 90 }^{ o }-{ 23 }^{ o } \right) } } -\frac { \tan { { 40 }^{ o } } }{ \tan { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } } -\sin { { 90 }^{ o } }$

$\displaystyle =\frac { 2\cos { { 67 }^{ o } } }{ \cos { { 67 }^{ o } } } -\frac { \tan { { 40 }^{ o } } }{ \tan { { 40 }^{ o } } } -\sin { { 90 }^{ o } }$

$\displaystyle =2-1-1=0$

The value of

$\displaystyle \sin { \theta } \cos { \theta } -\frac { \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } \cos { \theta } }{ \sec { \left( { 90 }^{ o }-\theta \right) } } -\frac { \cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } \sin { \theta } }{ \text{cosec }\left( { 90 }^{ o }-\theta \right) }$ is :

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

Explanation

Let

$A = \displaystyle \sin { \theta } \cos { \theta } -\frac { \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } \cos { \theta } }{ \sec { \left( { 90 }^{ o }-\theta \right) } } -\frac { \cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } \sin { \theta } }{ \text{cosec }\left( { 90 }^{ o }-\theta \right) }$
Using the identities

$\cos({90}^{o}-\theta) =\sin{\theta}, \sin( {90}^{o}-\theta) =\cos{\theta}, \sec({90}^{o}-\theta)=\text{cosec }\theta, \text{cosec}({90}^{o}-\theta) =\sec{\theta}$ , we get

$A =\sin { \theta } \cos { \theta } -\dfrac { \sin { \theta } \sin { \theta } \cos { \theta } }{ \text{cosec }\theta } -\dfrac { \cos { \theta } \cos { \theta } \sin { \theta } }{ \sec { \theta } }$

$\displaystyle =\sin { \theta } \cos { \theta }- { \sin }^{ 3 }\theta \cos { \theta } -{ cos }^{ 3 }\theta \sin { \theta }$ .....

$\displaystyle \left[ \because \frac { 1 }{ \text{cosec }\theta } =\sin { \theta}, \frac { 1 }{ \sec { \theta } } =\cos { \theta } \right]$

$\displaystyle =\sin { \theta } \cos { \theta } -\sin { \theta } \cos { \theta } \left( { \sin }^{ 2 }\theta +{ cos }^{ 2 }\theta \right)$

$\displaystyle =\sin { \theta } \cos { \theta } -\sin { \theta } \cos { \theta } \left( 1 \right)$

$\displaystyle =\sin { \theta } \cos { \theta } -\sin { \theta } \cos { \theta } =0$

$\displaystyle \sec 2A=\text{cosec } \left ( A-42^{\circ} \right )$ where

$2A$ is acute angle, then value of

$A$ is

0%
$\displaystyle 44^{\circ}$
0%
$\displaystyle 22^{\circ}$
0%
$\displaystyle 21^{\circ}$
0%
$\displaystyle 66^{\circ}$

Explanation

Given,

$\displaystyle \sec 2A=cosec \left ( A-42^{\circ} \right )$

$\displaystyle \left [ \because \sec \theta = cosec \left ( 90^{\circ}-\theta \right ) \right ]$

$\displaystyle \Rightarrow cosec \left ( 90^{\circ}-2A \right )= cosec \left ( A-42^{\circ} \right )$

$\displaystyle\Rightarrow \left ( 90^{\circ}-2A \right )=\left ( A-42^{\circ} \right )$

$\displaystyle \Rightarrow 90^{\circ}+42^{\circ}=A+2A$

$\displaystyle \Rightarrow 132^{\circ}=3A$

$\displaystyle \Rightarrow A=44^{\circ}$

Find the value of

$4\cos^2{60^0}+4\tan^2{45^0}-cosec^2\ {30^0}$

0%
$0$
0%
$2$
0%
$1$
0%
$\displaystyle\frac{1}{2}$

Explanation

Substituting the corresponding values we get

$4 \cos^2 60^0 + 4 \tan^2 45^0 - cosec ^ 2 \ 30^0 = 4 \left(\dfrac {1}{2}\right)^2 + 4(1)^2 - 2^2 = 1 + 4 - 4 = 1$

$\displaystyle \sin \theta =\dfrac12$ and

$\displaystyle 0^{\circ}< \theta < 90^{\circ}$ then

$\displaystyle \cos 2\theta =$ _____

0%
$0$
0%
$1$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{\sqrt{3}}{2}$

Explanation

Given,

$\sin { \theta } = \dfrac {1}{2}$

$=> \theta = {30}^{0}$ (Since,

$\sin { {30}^{0} } = \dfrac {1}{2}$ and

${0}^{0}<\theta< {90}^{0})$

So,

$\cos { 2\theta } = \cos ({60}^{0}) = \dfrac {1}{2}$

If co

$\displaystyle \sec \left ( 20^{\circ} + x \right )=\sec \left ( 50^{\circ}+x \right )$ the value of x is

0%
$\displaystyle 10^{\circ}$
0%
$\displaystyle 20^{\circ}$
0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 40^{\circ}$

Explanation

We know that

$cosec (A) = sec(90-A)$

This means, if

${20}^{0} + x = A$ ,
then

${50}^{0} + x = {90}^{0} - ({20}^{0} + x)$

$=> {50}^{0} + x = {90}^{0} - {20}^{0}-x$

$=> 2 x = {20}^{0}$

$=> x = {10}^{0}$

The value of

$\displaystyle 4\left ( \sin ^{4}30^{\circ}+\cos ^{4}30^{\circ} \right )-3\left ( \cos ^{2}45^{\circ}+\sin ^{2}90^{\circ} \right )$ is:

0%
$-\dfrac12$
0%
$-2$
0%
$2$
0%
$\dfrac12$

Explanation

Given:

$4\left( \sin ^{ 4 }{ 30° } +\cos ^{ 4 }{ 30° } \right) -3\left( \cos ^{ 2 }{ 45° } +\sin ^{ 2 }{ 90° } \right)$

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

$=4\left( \dfrac { 1 }{ 16 } +\dfrac { 9 }{ 16 } \right) -3\left( \dfrac { 1 }{ 2 } +1 \right)$

$=\left( 4\times \dfrac { 10 }{ 16 } \right) -\left( 3\times \dfrac { 3 }{ 2 } \right)$

$=\dfrac { 5 }{ 2 } -\dfrac { 9 }{ 2 }$

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

$=-2$

Hence, the answer is

$-2.$

$\displaystyle \sin^{2} 85^{\circ} + \sin ^{2}5^{\circ} =$ ______

0%
$0$
0%
$1$
0%
$\dfrac12$
0%
$\dfrac23$

Explanation

Given,

$\sin ^{ 2 }{ { 85 }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } }$

We know that

$\sin ^{ 2 }{ { A }^{ o } } = \cos ^{ 2 }{ { (90-A) }^{ o } }$

$\Rightarrow \sin ^{ 2 }{ { 85 }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } = \cos ^{ 2 }{ {(90-85) }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } = \cos ^{ 2 }{{ 5 }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } = 1$

(Since,

$\cos ^{ 2 }{ {A }^{ o } } + \sin ^{ 2 }{ { A }^{ o } } = 1$ )

$\displaystyle \sin ^{4}\theta +\cos ^{4}\theta =\dfrac{1}{2}$ then the value of

$\displaystyle \sin \theta \cos \theta$ is

0%
$\displaystyle \pm \frac{1}{8}$
0%
$\displaystyle \pm \frac{1}{4}$
0%
$\displaystyle \pm 1$
0%
$\displaystyle \pm \frac{1}{2}$

Explanation

We know that

${ a }^{ 4 }+{ b }^{ 4 }=\quad \left( { a }^{ 2 }+b^{ 2 } \right) ^{ 2 }-2a^{ 2 }{ b }^{ 2 }$

So,

$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ 0 } =\quad \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) ^{ 2 }-2(\sin ^{ 2 }{ \theta } )^{ 2 }{ (\cos ^{ 2 }{ \theta } ) }^{ 2 }$

But

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

So,

$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta }= 1^{2} -2(\sin \theta \cos \theta) ^ {2}$

$\Rightarrow \dfrac {1}{2} = 1 - 2(\sin \theta \cos \theta) ^ {2}$

$\Rightarrow 2(\sin \theta \cos \theta) ^ {2}= \dfrac {1}{2}$

$\Rightarrow (\sin \theta \cos \theta) ^ {2}= \dfrac {1}{4}$

$\Rightarrow \sin \theta \cos \theta = \pm \dfrac {1}{2}$

$\displaystyle \sqrt{3}\tan \theta =3\sin \theta$ then the value of

$\displaystyle \sin ^{2}\theta -\cos ^{2}\theta$ is

0%
2
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{3}$
0%
3

Explanation

Given

$\sqrt{3}\tan\theta=3\sin \theta$

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

$\sqrt 3\dfrac{\sin \theta}{\cos \theta}-3\sin \theta=0$

$\sqrt3\sin \theta \big(\dfrac{1}{\cos \theta}-\sqrt3\big)=0$

Now either

$\sin \theta=0$ or

$\cos \theta=\dfrac{1}{\sqrt3}$

But

$\sin \theta \neq 0$

Hence

$\cos \theta =\dfrac{1}{\sqrt3}$

Now u\sin g

$\sin ^2\theta+\cos ^2\theta=1$ ,we will get

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

Hence

$\sin ^2\theta-\cos ^2\theta=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}$

$\displaystyle \sin \Theta +\text{cosec }\Theta =2$ find the value of

$\displaystyle \cot \Theta +\cos \Theta$

0%
$0$
0%
$\dfrac12$
0%
$1$
0%
$2$

Explanation

Given,

$\sin \Theta + \text{cosec } \Theta = 2$

This is possible only when

$\Theta = 90^o$ as

$\sin 90^0 = \text{cosec } 90^o = 1$

So,

$\cot \Theta + \cos \Theta = \cot 90^o + \cos 90^ 0 = 0 + 0 = 0$

$\displaystyle X=\tan 1^{0}+\tan 2^{0}+........+\tan 45^{0}$ and

$\displaystyle y= -(\cot 46^{0}+\cot 47^{0}+.......+\cot 89^{0})$ then find the value of

$(x + y)$ .

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

Explanation

Since

$\tan (90^o - x) = \cot 90^0$ we have

$x + y = \tan 1^o + \tan 2^o +....... + \tan 45^0 - (\cot 46^0 + \cot 47^0 + ... + \cot 89 ^0 )$

$=>x+y= \tan 1^o + \tan 2^o +.... + \tan 45^0 - (\cot (90^0 - 44^0) + \cot(90^0-43^0) + .. +\cot (90^0 - 1 ^0 )$

$=>x+y= \tan 1^o + \tan 2^o +.... + \tan 45^0 - (\tan 44^0 + \tan43^0 + .. + \tan 1 ^0 )$

$=> x+y = 0 + 0 + .. + \tan 45^0 = 1$

$\displaystyle \sec ^{4}\theta -\sec ^{2}\theta$ in terms of

$\displaystyle \tan \theta$ is ___

0%
$\displaystyle \tan ^{4}\theta -\tan ^{2}\theta$
0%
$\displaystyle \tan ^{4}\theta +\tan ^{2}\theta$
0%
$\displaystyle \tan ^{2}\theta -\tan ^{4}\theta$
0%
None of these

Explanation

$\displaystyle \sec ^{4}\theta -\sec ^{2}\theta=(\sec^2\theta)^2-\sec^2\theta=(1+\tan ^2\theta)^2-(1+\tan ^2\theta)=1+\tan ^4\theta+2\tan ^2\theta-1-\tan ^2\theta\\=\tan ^4\theta+\tan ^2\theta$

Therefore, Answer is

$\tan ^4\theta+\tan ^2\theta$

$\displaystyle \sin \left ( A+B \right )=\dfrac{\sqrt{3}}{2}$ and

$\displaystyle \cot \left ( A-B \right )=\sqrt{3}$ then find the value of

$B$ .

0%
$\displaystyle 15^{\circ}$
0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 10^{\circ}$
0%
$\displaystyle 20^{\circ}$

Explanation

Given,

$\sin(A+B) = \dfrac {\sqrt{3}}{2}$

$\Rightarrow A +B = 60^o$ -(1)

And

$\cot(A-B) = \sqrt {3}$

$\Rightarrow A-B = 30^o$ ---- (2)

Subtracting eqn 2 from 1

$\Rightarrow 2B= 30^0$

$\Rightarrow B= 15^0$

$\displaystyle \cos \Theta _{1}+\cos \Theta _{2}+\cos \Theta _{3}=3,$ find

$\displaystyle \sin \Theta _{1}+\sin \Theta _{2}+\sin \Theta _{3}.$

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

Explanation

Maximum value of a cosine of an angle is

$1$

Hence, when sum of three cosines

$= 3$ , this means each cosine

$= 1$
And each angle

$= 0^{o}$

So

$\sin 0 + \sin 0 + \sin 0 = 0$

The simplified value of

$\displaystyle \text{cosec} ^{2}\alpha \left ( 1+\frac{1}{\sec \alpha } \right )\left (1-\frac{1}{\sec \alpha } \right )$ is _____

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

Explanation

We know that

$\text{cosec}\, \alpha = \dfrac {1}{\sin \alpha}$ and

$\sec \alpha = \dfrac {1}{\cos \alpha}$

So,

$\text{cosec}\, ^2 \alpha \left(1 + \dfrac {1}{\sec \alpha}\right)\left(1 - \dfrac {1}{\sec \alpha}\right) = \dfrac {1}{\sin ^2 \alpha}\left(1+\cos \alpha\right)\left(1-\cos \alpha\right)$

$= \dfrac {1}{\sin ^2 \alpha}\left(1-\cos ^2\alpha\right) = \dfrac {1}{\sin ^2 \alpha}\left(\sin ^2\alpha\right) = 1$

Value of

$\left(\tan{30^0}\csc{60^0} + \tan{60^0}\sec{30^0}\right)$ is

0%
$2\dfrac{1}{3}$
0%
$2\dfrac{2}{3}$
0%
$3\dfrac{1}{3}$
0%
$\dfrac{7}{4}$

Explanation

Taking

$\left( \tan { 30^0 } \csc { 60^0+\tan { 60^0 } \sec { 30^0 } } \right)$

Substituting the values of tan

$30^0$ ,csc

$60^0$ , tan

$60^0$ ,sec

$30^0$

$=\left( \dfrac { 1 }{ \sqrt { 3 } } \times \dfrac { 2 }{ \sqrt { 3 } } \right) +\left( \sqrt { 3 } \times \dfrac { 2 }{ \sqrt { 3 } } \right)$

$=\dfrac { 2 }{ 3 } +2$ taking L.CM we get

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

The value of

$\displaystyle \sin ^{2}5^{\circ}+ \sin ^{2}10^{\circ}+\sin ^{2}15^{\circ}......+\sin ^{2}90^{\circ}$ is equal to

0%
$\dfrac{17}2$
0%
$\dfrac{19}2$
0%
$\dfrac{21}2$
0%
$\dfrac{23}2$

Explanation

$\displaystyle \sin ^{2}5^{\circ}+ \sin ^{2}10^{\circ}+\sin ^{2}15^{\circ}......+\sin ^{2}90^{\circ}$

$=(\sin^25^o+\sin^285^o)+(\sin^210^o+\sin^2{80^o})+...........+(\sin^245^o+\sin^245^o)+\sin^290^o-\sin^245^o$

$=(\sin^25^o+\cos^25^o)+(\sin^210^o+\cos^2{10^o})+...........+(\sin^245^o+\cos^245^o)+\sin^290^o-\sin^245^o$

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

$=10-\dfrac{1}{2}=\dfrac{19}{2}$

$\displaystyle\tan \theta +\cot \theta =2$ find the value of

$\displaystyle \tan ^{1025}\theta +\cot ^{1025}\theta$

0%
$0$
0%
$1$
0%
$2$
0%
$\dfrac12$

Explanation

Given

$\tan\theta+\cot\theta=2$

$\Rightarrow \tan\theta+\dfrac{1}{\tan\theta}=2$

$\Rightarrow \tan^2\theta+1=2\tan\theta$

$\Rightarrow \tan^2\theta-2\tan\theta+1=0$

$\Rightarrow (\tan\theta-1)^2=0$

$\Rightarrow \tan\theta=1\therefore \cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{1}=1$

$\therefore \tan^{1025}\theta+\cot^{1025}\theta=1^{1025}+1^{1025}=1+1=2$

$\displaystyle \tan \theta -\cot \theta =7$ find the value of

$\displaystyle \tan ^{3}\theta -\cot^{3} \theta$

0%
$284$
0%
$296$
0%
$345$
0%
$364$

Explanation

$\displaystyle \tan ^{3}\theta -\cot^{3} \theta=( \tan\theta- \cot\theta)(\tan^2\theta+\cot^2\theta+\tan\theta \cot\theta)\\=( \tan\theta- \cot\theta)((\tan\theta-\cot \theta)^2+2\tan\theta cot \theta+\tan \theta \cot \theta)\\=7(7^2+3\tan\theta \cot \theta)=7(49+3)\\=7\times52=364$

Therefore, Answer is

$364$

$\displaystyle 3\cos ^{2}A=\cos 60^{\circ}+\sin ^{2}45^{\circ}$ then find the value of

$\displaystyle \sec ^{2}A$ .

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

Explanation

$3\cos^2 A = \cos 60^0 + \sin^2 45$

$\Rightarrow 3\cos^2 A = \dfrac {1}{2} + (\dfrac {1}{\sqrt {2}})^2$

$\Rightarrow 3\cos^2 A = \dfrac {1}{2} + \dfrac {1}{2} = 1$

$\Rightarrow \cos^2 A= \dfrac {1}{3}$

Or

$\sec^2 A = \dfrac {1}{\cos^2 A} = 3$

The simplified value of

$\displaystyle \left ( \frac{1-\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{1+\sin \alpha } \right )\left ( \sec \alpha +\frac{1}{\cot \alpha } \right )$ is _____

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

Explanation

$\left( \dfrac { 1-\sin \alpha }{ \cos \alpha } +\dfrac { \cos \alpha }{ 1+\sin \alpha } \right) \left( \sec \alpha \quad +\quad \dfrac { 1 }{ \cot \alpha } \right) \\= \left( \dfrac { (1-\sin \alpha )(1+\sin \alpha )+\cos ^ 2 \alpha \quad }{ (1+\sin \alpha )\cos \alpha } \right) \left( \dfrac { 1 }{ \cos \alpha } +\quad \dfrac { \sin \alpha }{ \cos \alpha } \right)$
$\\ = \left( \dfrac { (1-\sin ^{ 2 }\alpha +\cos ^{ 2 }\alpha \quad }{ (1+\sin \alpha )\cos \alpha } \right) \left( \dfrac { 1+\sin \alpha }{ \cos \alpha } \right) \\= \left( \dfrac { \cos ^{ 2 }\alpha +\cos ^{ 2 }\alpha \quad }{ \cos \alpha } \right) \left( \dfrac { 1 }{ \cos \alpha } \right) \\= 2$

$\displaystyle \text{cosec}\, \theta -\cot \theta =2,$ then find the value of

$\displaystyle \text{cosec} ^{2}\theta +\cot^{2} \theta.$

0%
$\displaystyle \frac{8}{15}$
0%
$\displaystyle \frac{15}{8}$
0%
$\displaystyle \frac{8}{17}$
0%
$\displaystyle \frac{17}{8}$

Explanation

$cosec\;\theta-\cot\theta=2 \longrightarrow \left( 1 \right)$

Now,

$\left( cosec^{ 2 }\theta -\cot ^{ 2 }{ \theta } \right) =1$

$\Rightarrow \left( cosec\;\theta -\cot { \theta } \right) \left( cosec\;\theta +\cot { \theta } \right) =1$

$\Rightarrow 2\left( cosec\;\theta +\cot { \theta } \right) =1$

$\Rightarrow \left( cosec\;\theta +\cot { \theta } \right) =\dfrac { 1 }{ 2 } \longrightarrow \left( 2 \right)$

Adding

$\left( 1 \right) \;\&\;\left( 2 \right) ,$ we get

$\Rightarrow 2\;cosec\;\theta =2+\dfrac { 1 }{ 2 } =\dfrac { 5 }{ 2 }$

$\Rightarrow cosec\;\theta =\dfrac { 5 }{ 4 }$

$\Rightarrow cosec^{ 2 }\theta =\dfrac { 25 }{ 16 }$

Again

$\left( 2 \right) \;-\;\left( 1 \right) ,$ we get

$\Rightarrow 2\cot { \theta } =\dfrac { 1 }{ 2 } -2=-\dfrac { 3 }{ 2 }$

$\Rightarrow \cot { \theta } =-\dfrac { 3 }{ 4 }$

$\Rightarrow \cot ^{ 2 }{ \theta } =\dfrac { 9 }{ 16 }$

$\Rightarrow cosec^{ 2 }\theta +\cot ^{ 2 }{ \theta } =\dfrac { 34 }{ 16 } =\dfrac { 17 }{ 8 }$

In a right angled

$\triangle ABD$ ,

$\angle B = 60^o$ and

$\angle A = 30^o$ .

Then,

$\cos{60^o} =$

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

Explanation

Consider an equilateral

$\triangle ABC$ with each side

$= 2a$ .

Each angle of

$\triangle ABC$ is

$60^o$ .

From

$A$ , draw

$AD \perp BC$ , then

$BD = DC = a$ .

Also,

$\angle ADB = 90^o, \angle BAD = 30^o$ .

From right angled

$\triangle ADB$ , we have

$AD = \sqrt{{AB}^{2}-{BD}^{2}} = \sqrt{{\left(2a\right)}^{2}-{\left(a\right)}^{2}}$

$= \sqrt{4{a}^{2}-{a}^{2}} = \sqrt{3{a}^{2}} = \sqrt{3}a$ .

Now, in right angled

$\triangle ADB$ , base

$BD = a, \, perpendicular = AD = \sqrt{3}a$

and hypotenuse

$= AB = 2a$

$\cos{60^o} = \dfrac{BD}{AB} = \dfrac{a}{2a} = \dfrac{1}{2}$

So,

$\dfrac{1}{2}$ is correct.

In a right angled

$\triangle ABD$ , in which

$\angle B=60^o$ and

$\angle A=30^o$ .

Then

$\tan { 30^o }$ is equal to

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

Explanation

Consider an equilateral

$\triangle ABC$ , with each side

$= 2a$ .

So, each angle is

$60^o$ . From

$A$ , draw

$AD \perp BC$ .

So,

$BD=DC=a, \angle ADB=90^o, \angle BAD=30^o$ .

So, for right angled

$\triangle ADB$ , we have

$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$ .

$\tan { 30 } =\dfrac { BD }{ AD } =\dfrac { a }{ \sqrt { 3 } a } =\dfrac { 1 }{ \sqrt { 3 } }$ .

So,

$\dfrac { 1 }{ \sqrt { 3 } }$ is correct.

In a right angled

$\triangle ABD$ ,

$\angle B = 60^o$ and

$\angle A = 30^o$ , then

$\sin{60^o} =$

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

Explanation

Consider an equilateral triangle with each side

$= 2a$ .

So each angle of

$\triangle ABC$ is

$60^o$ .

From

$A$ draw

$AD \perp BC$ then

$BD = DC = a$ .
Also,

$\angle ADB = 90^o$ and

$\angle BAD = 30^o$

From right angled triangle

$\triangle ADB$ , we have

$AD = \sqrt{{AB}^{2}-{BD}^{2}} = \sqrt{{\left(2a\right)}^{2}-{\left(a\right)}^{2}}$

$= \sqrt{4{a}^{2} - {a}^{2}} = \sqrt{3{a}^{2}} = \sqrt{3}a$

Now, in right angled

$\triangle ADB$ , base

$BD = a, AD = \sqrt{3}a$

and hypotenuse

$= AB = 2a$

$\sin{60^o} = \dfrac{AD}{AB} = \dfrac{\sqrt{3} \times a}{2 \times a} = \dfrac{\sqrt{3}}{2}$

So,

$\dfrac{\sqrt{3}}{2}$ is correct.

In a right angled triangle,

$\angle A = \theta$ is an acute angle.

Then,

$\cos{\theta} \times \sec{\theta}$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
None of these

Explanation

$\cos{\theta} = \dfrac{x}{r}$ and

$\sec{\theta} = \dfrac{r}{x}$

$\therefore \cos{\theta} \times \sec{\theta} = \dfrac{x}{r} \times \dfrac{r}{x} = 1$

In a right angle

$\triangle ABD$ , in which

$\angle B = 60^o$ and

$\angle A = 30^o$ .

Then,

$\tan{60^o}$ is equal to

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

Explanation

Consider an equilateral

$\triangle ABC$ with each side

$= 2a$ .
So each angle is

$60$ .
From

$A$ , draw

$\perp AD$ to

$BC$ ,

$BD = DC = a$ and

$\angle ADB = 90^o, \angle BAD = 30^o$ .
So, for

$\triangle ADB$ , we have

$AD = \sqrt{{AB}^{2} - {BD}^{2}}$

$= \sqrt{{\left(2a\right)}^{2}-{a}^{2}} = \sqrt{4{a}^{2}-{a}^{2}} = \sqrt{3{a}^{2}} = \sqrt{3}a$ .

$\therefore \tan 60^o=\dfrac{AD}{BD}=\dfrac{\sqrt 3 a}{a}=\sqrt 3$

In a right angle

$\triangle ABD, \angle B=60^o$ and

$\angle A=30^o$ .

Then

$\csc{60^o}$ is equal to:

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

Explanation

Consider an equilateral

$\triangle ABC$ with each side

$= 2a$ .
Each angle is

$60^o$ .
From

$A$ , draw

$AD \perp BC$ , then

$BD = DC = a$ .
Also,

$\angle ADB=90^o, \angle BAD=30^o$ .
From right angle

$\triangle ADB$ , we have

$AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } =\sqrt { { \left( 2a \right) }^{ 2 }-{ a }^{ 2 } }$

$=\sqrt { { 4a }^{ 2 }-{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } =\sqrt { 3 } a$

$\csc { 60 } =\dfrac { 1 }{ \sin { 60 } } =\dfrac { 1 }{ \dfrac { \sqrt { 3 } }{ 2 } } =\dfrac { 2 }{ \sqrt { 3 } }$
So

$\dfrac { 2 }{ \sqrt { 3 } }$ is correct.

If in a rectangular, the angle between a diagonal and a side is 30

$^o$ and the length of the diagonal is

$8$ cm, then the area of the rectangle is _____ .

0%
$16$ $\sqrt3$ cm $^2$
0%
$16$ cm $^2$
0%
$\sqrt3$ cm $^2$
0%
16/ $\sqrt3$ cm $^2$

Explanation

Referring figure, let ABCD be a rectangular in which

$\angle$ BAC = 30

$^o$ and AC = 8 cm

Then, cos30

$^o$ =

$\displaystyle{\frac{AB}{AC} \Rightarrow \frac{\sqrt3}{2} = \frac{AB}{8} \Rightarrow}$ AB = 4

$\sqrt3$ cm

and. sin 30

$^o$ =

$\displaystyle{\frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{BC}{8} \Rightarrow}$ BC = 4 cm

$\therefore$ Area of rectangular = AB x BC
=4

$\sqrt3$ x 4
= 16

$\sqrt3$ cm

$^2$

So, option A is correct.

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 8 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 8 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.