MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 10

$\csc (90 - \theta) =$ is equivalent to

0%
$\sec \theta$
0%
$\sin \theta$
0%
$\cos \theta$
0%
None of these

Explanation

In the figure,

$cosec (90 - \theta) = \dfrac {r}{x}$ and

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

So,

$cosec (90 - \theta) = \sec \theta$

Hence,

$\sec \theta$ is correct.

Solve:

$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec } 70^{\circ}$

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

Explanation

Given:

$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec} 70^{\circ}$

$= \sec 70^{\circ} \sin (90^{\circ} - 70^{\circ}) + \cos 20^{\circ} \text{cosec} (90^{\circ} - 20^{\circ})$

$= \sec 70^{\circ} \cos 70^{\circ} + \cos 20^{\circ} \sec 20^{\circ}$

$[\because \sin(90^{\circ}-\theta)=\cos \theta$ and

$\text{cosec}(90^{\circ}-\theta)=\sec \theta]$

$= 1 + 1$

$[\because ~\sec \theta.\cos \theta=1]$

$= 2$

So,

$2$ is correct.

$A$ and

$B$ are complementary angles, then

$\sin A \times \sec B =$

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

Explanation

Given that both the angles are complementary i.e.,

$A+B=90^{\circ}$

So,

$A=90^{\circ}-B$

We have :

$=\sin A\times \sec B$

$=\sin(90^o-B)\times \sec B$

$= \cos B\times \sec B$ [using

$\cos \theta = \sin (90^o - \theta)$ ]

$=\cos B \times \dfrac{1}{\cos B}$

$=1$

$\tan 5^{\circ} \tan 25^{\circ} \tan 30^{\circ} \tan 65^{\circ} \tan 85^{\circ} =$

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

Explanation

$(\tan 5^{\circ} \tan 85^{\circ}) (\tan 25^{\circ} \tan 65^{\circ}) \tan 30^{\circ}$

$= [\tan 5^{\circ} \tan (90^{\circ} - 5^{\circ})] [\tan 25^{\circ} (\tan 90^{\circ} - 25^{\circ})] \times \tan 30^{\circ}$

$= \tan 5^{\circ} \cot 5^{\circ} \tan 25^{\circ} \cot 25^{\circ} \times \dfrac {1}{\sqrt {3}}$

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

$\cot (90^{\circ} - \theta)$ is equivalent to

0%
$\cot \theta$
0%
$\cos \theta$
0%
$\tan \theta$
0%
$-\tan \theta$

Explanation

we know

$\cos (90 - \theta)= \tan \theta$

So,

$\tan \theta$ is correct.

$\text{cosec } (75^{\circ} + \theta) - \sec (15^{\circ} - \theta) =$

0%
$2\sec \theta$
0%
$2 cosec \theta$
0%
$0$
0%
$1$

Explanation

Given,

$\csc { \left( { 75^o } +\theta \right) } -\sec { \left( { 15^o } -\theta \right) }$

$=\csc { \left( { 75^o } +\theta \right) } -\sec { \left[ { { 90^o } -\left( 75^o+\theta \right) } \right] }$

$=\csc { \left( { 75^o } +\theta \right) } -\csc { \left( { 75^o } +\theta \right) }$

$=0$

$0$ will be the answer

$\tan 2A = \cot (A - 21^{\circ})$ , where

$2A$ is an acute angle, then

$\angle A =$ ____

0%
$24^{\circ}$
0%
$27^{\circ}$
0%
$35^{\circ}$
0%
$37^{\circ}$

Explanation

Given,

$\tan 2A = \cot (A - 21^{\circ})$

Therefore,

$\cot (90^{\circ} - 2A) = \cot (A - 21^{\circ})$

$\Rightarrow 90^{\circ} - 2A = A - 21^{\circ}$

$\Rightarrow 90^{\circ} + 21^{\circ} = 2A + A$

$\Rightarrow 111^{\circ} = 3A$

$\Rightarrow A = 37^{\circ}$
So,

$\angle A=37^{\circ}$ is correct.

$\sin^{2} 20^{\circ} + \sin^{2} 70^\circ=$ ?

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

Explanation

$\sin^{2} 20^{\circ} = \cos^{2} (90^{\circ} - 20^{\circ}) = \cos^{2} 70^\circ$

Substituting

$\sin^{2} 20^{\circ} = \cos^{2} 70^\circ$

$\sin^220^o+\sin^270^o=\cos^{2} 70^{\circ} + \sin^{2} 70^{\circ} = 1$

$[\because \sin^2\theta+\cos^2\theta=1]$

Evaluate:

$\dfrac {\tan 35^{\circ}}{\cot 55^{\circ}} + \dfrac {\cot 78^{\circ}}{\tan 12^{\circ}} =$

0%
$0$
0%
$1$
0%
$2$
0%
None of these

Explanation

Given expression is

$\dfrac { \tan { { 35^o } } }{ \cot { { 55^o } } } +\dfrac { \cot { { 78^o } } }{ \tan { { 12^o } } }$

We can rewrite the expression as:

$\dfrac { \tan { { 35^o } } }{ \cot { \left( { 90^o- } { 35^o } \right) } } +\dfrac { \cot { { 78^o } } }{ \tan { \left( { 90^o- } { 78 } \right) } }$

We know that,

$\cot { \left( 90^o-\theta \right) } =\tan { \theta }$ and

$\tan { \left( 90^o-\theta \right) } =\cot { \theta }$

$\therefore \$ The expression becomes,

$\dfrac { \tan { { 35^o } } }{ \tan { { 35^o } } } +\dfrac { \cot { { 78^o } } }{ \cot { { 78^o } } }$

$\Rightarrow 1+1$

$\Rightarrow 2$

Hence,

$\dfrac { \tan { { 35^o } } }{ \cot { { 55^o } } } +\dfrac { \cot { { 78^o } } }{ \tan { { 12^o } } } =2$

$\sin 15^{\circ} = \cos (n\times15^{\circ})$ , then

$n = .....$

0%
$1$
0%
$2$
0%
$5$
0%
$0$

Explanation

As we know that

$\cos (90^{\circ}-\theta)=\sin \theta$

$\sin 15^{\circ}=\cos (90^{\circ}-15^{\circ})=\cos 75^{\circ}=\cos (5\times 15^{\circ})$

$n=5$

$\cot 1^{\circ} \cot 2^{\circ} ...... \cot 89^{\circ} =$

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

Explanation

$\cot 1^{\circ} \cot 2^{\circ} ...... \cot 89^{\circ}$

$=(\cot 1^o\cdot \cot 89^o)(\cot 2^o\cdot \cot 88^o)...............(\cot 44^o\cdot \cot 46^o)(\cot 45^o)$

$=(\cot 1^o\cdot \tan 1^o)(\cot 2^o\cdot \tan 2^o)...............(\cot 44^o\cdot \tan 44^o)(\cot 45^o)$ , since

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

$=(1)(1).......(1)(1)=1$ , since

$\tan\theta\cdot \cot\theta=1$

Without trigonometric table, evaluate

$\dfrac {\cot 63^{\circ}}{\tan 27^{\circ}}$

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

Explanation

Taking

$\dfrac {\cot 63^{\circ}}{\tan 27^{\circ}}$

$\dfrac {\cot (90-27)^{\circ}}{\tan 27^{\circ}}$

We know,

$\cot { \left( 90^0-\theta \right) } =\tan { \theta }$

$\dfrac {\tan27^{\circ}}{\tan 27^{\circ}}$

$=1$

Find the value of

$\tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ}$

0%
$\dfrac {1}{16}$
0%
$0$
0%
$1$
0%
None of these

$\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} ... \tan 89^{\circ} =$

0%
$1$
0%
$-1$
0%
$0$
0%
None of above

Explanation

$(\tan 1^{\circ} \tan 89^{\circ}) (\tan 2^{\circ} \tan 88^{\circ}) ..... (\tan 44^{\circ} \tan 46^{\circ}) \tan 45^{\circ}$ , group pair of terms

$= \left \{\tan 1^{\circ} \tan (90^{\circ} - 1^{\circ})\right \} \left \{\tan 2^{\circ} \tan (90^{\circ} - 2^{\circ})\right \} ...... \left \{\tan 44^{\circ} \tan (90^{\circ} - 44^{\circ})\right \}\tan 45^{\circ}$

$= (\tan 1^{\circ} \cot 1^{\circ})(\tan 2^{\circ} \cot 2^{\circ}) ..... (\tan 44^{\circ} \cot 44^{\circ}) \tan 45^{\circ}$

$= (1 \times 1 ..... 1)\times 1 = 1$

Note that

$\tan\theta\times \cot\theta=1$

Without trigonometric table, evaluate

$\dfrac {\sec 41^o}{\text{cosec } 49^o}$

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

Explanation

Given

$\dfrac { \sec { { 41^o } } }{ \csc { { 49^o } } }$

$=\dfrac { \sec { \left( { 90^o- } { 49^o } \right) } }{ \csc { { 49^o } } }$

$=\dfrac { \csc { { 49^o } } }{ \csc { { 49^o } } }$ as we know

$\sec { \left( 90^o-\theta \right) } =\csc { \theta }$

$=1$

Hence

$1$ is the answer

Find the value of

$\cos^2 \theta (1 + \tan^2 \theta) + \sin^2 \theta (1 + \cot^2 \theta)$ .

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

Explanation

we know that ,

$1+\tan ^2\theta=\sec ^2\theta\\1+\cot ^2\theta=\text{cosec} ^2\theta$

Now,

$\cos^2\theta(1+\tan ^2\theta )+\sin^2\theta(1+\cot ^2\theta)$

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

Therefore, Answer is

$2$

$\sin 81^{\circ} + \tan 81^{\circ}$ , when expressed in terms of angles between

$0^{\circ}$ and

$45^{\circ}$ , becomes

0%
$\sin 9^{\circ} + \cos 9^{\circ}$
0%
$\cos 9^{\circ} + \tan 9^{\circ}$
0%
$\sin 9^{\circ} + \tan 9^{\circ}$
0%
$\cos 9^{\circ} + \cot 9^{\circ}$

Explanation

We know that,

$\sin A = \cos(90˚-A)$

$\tan A = \cot(90˚-A)$

$\sec A = \text{cosec }(90˚-A)$

Using these, we get

$\sin 81^{\circ} = \sin (90^{\circ} - 9^{\circ}) = \cos 9^{\circ}$

$\tan 81^{\circ} = \tan (90˚ - 9) = \cot 9^{\circ}$

$\therefore \sin 80^{\circ} + \tan 81^{\circ} = \cos 9^{\circ} + \cot 9^{\circ}$

Find the value of :

$\dfrac {\cos 38^{\circ} \csc 52^{\circ}}{\tan 18^{\circ} \tan 35^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}} =$

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

Explanation

Given:

$\dfrac { \cos { { 38^o } \csc { { 52^o } } } }{ \tan { { 18^o } } \tan { { 35^o } \tan { { 60^o } } \tan { { 72^o } } \tan { { 55^o } } } }$

$=\dfrac { \cos { { 38^o } } \csc { \left( { 90^o- } { 38^o } \right) } }{ \left( \tan { { 18^o\times } } \tan { { 72^o } } \right) \times \left( \tan { { 35^o\times } } \tan { { 55^o } } \right) \times \tan { { 60^o } } }$

$=\dfrac { \cos { { 38^o } } \sec { { 38^o } } }{ \tan { { 18^o } } \tan { \left( { 90^o-18^o } \right) \tan { { 35^o } } \tan { \left( { 90^o- } { 35^o } \right) \tan { { 60^o } } } } }$

( we know that

$\tan { \left( 90^o-\theta \right) } =\cot { \theta }$ and

$\cot { \theta } =\dfrac { 1 }{ \tan { \theta } }$ and

$\sec { \theta =\dfrac { 1 }{ \cos { \theta } } }$ and

$\tan { { 60^o=\sqrt { 3 } } } )$

$\therefore \dfrac { \cos { { 38^o } \csc { { 52^o } } } }{ \tan { { 18^o } } \tan { { 35^o } \tan { { 60^o } } \tan { { 72^o } } \tan { { 55^o } } } }$

$=\dfrac { 1 }{ \left( \tan { { 18^o } } \cot { { 18^o } } \right) \left( \tan { { 35^o } } \cot { { 35^o } } \right) \sqrt { 3 } }$

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

Evaluate:

$\dfrac {\tan 30^{\circ}}{\cot 60^{\circ}}$

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

Explanation

Taking

$\dfrac {\tan 30^{\circ}}{\cot 60^{\circ}}$

$\dfrac {\tan 30^{\circ}}{\cot (90-30)^{\circ}}$

as we know,

$\cot { \left( 90^0-\theta \right) } =\tan { \theta }$

$\dfrac {\tan 30^{\circ}}{\tan30^{\circ}}$

$=1$

Find the name of the person who first produce a table for solving a triangle's length and angles.

0%
William Rowan Hamilton
0%
Hipparchus
0%
Euclid
0%
Issac Newton

Explanation

$\text {Hipparchus}$ gave the first table of chords analogus to modern table of sine values, and used them to solve trigonometric problems

What is the meaning of trigonometry in Greek language?

0%
Measurement
0%
Triangle Measure
0%
Angle Measure
0%
Degree Measure

Explanation

Trigonometry in the Greek language means triangle measure.

Trigono

$\Rightarrow$ triangle

$+$ metron

$\Rightarrow$ measure

So, option

$B$ is correct.

$A + B = 90^o$ , then ......

0%
$\sin A = \sin B$
0%
$\cos A = \cos B$
0%
$\tan A = \tan B$
0%
$\sec A = \text{cosec } B$

Explanation

$\sec A = \dfrac {r}{x}$ and

$\text{cosec } A = \dfrac {r}{x}$

So,

$\sec A = \text{ cosec } A$

$\dfrac {\cos (90 -\theta) \sec (90 - \theta)\tan \theta}{\text{cosec } (90 - \theta)\sin (90 - \theta) \cot (90 - \theta)} + \dfrac {\tan (90 - \theta)}{\cot \theta} = ......$

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

Explanation

As we know that,

$\sin(90^{o}-\theta)=\cos \theta$ ,

$\cos(90^o-\theta)=\sin \theta$ ,

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

$\cot(90^o-\theta)=\tan\theta$ ,

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

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

$\therefore\ \dfrac{\cos(90-\theta) \sec(90-\theta) \tan \theta}{\text{cosec}(90-\theta) \sin(90-\theta) \cot(90-\theta)} + \dfrac{\tan(90-\theta)}{\cot \theta}$

$\therefore\ \dfrac{\sin \theta \times \frac{1}{\cos(90-\theta)} \times \tan \theta}{\frac{1}{\sin(90-\theta)} \times \sin(90-\theta) \times \tan \theta} + \dfrac{\cot \theta}{\cot \theta}$

$=\dfrac{\sin \theta \times \frac{1}{\sin \theta}}{\frac{1}{\cos \theta} \times \cos \theta} + 1$

$=1+1=2$

Hence, option C is correct.

In the 5th century who created the table of chords with increasing 1 degree?

0%
Hipparchus
0%
William Rowan Hamilton
0%
Euclid
0%
Ptolemy

The value of

$\cot 1^{\circ} \cot 2^{\circ} .... \cot 89^{\circ}$ is .....

0%
$1$
0%
$0$
0%
$-1$
0%
None of above

Explanation

The given equation can be written as

$(\cot 1^{\circ} \cot 89^{\circ})(\cot 2^{\circ} \cot 88^{\circ}) ..... (\cot 44^{\circ} \cot 46^{\circ}) \cot 45^{\circ}$

$= (\cot 1^{\circ} \cot (90^{\circ} - 1^{\circ})) (\cot 2^{\circ} \cot (90^{\circ} - 2^{\circ})) ..... (\cot 44^{\circ} \cot (90^{\circ} - 44^{\circ}) ) \cot 45^{\circ}$

$= \cot 1^{\circ} \tan 1^{\circ} \cdot \cot 2^{\circ} \tan 2^{\circ} .... \cot 44^{\circ} \tan 44^{\circ} \times \cot 45^{\circ}$ (Since

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

$= 1\times 1 ..... 1\times 1$ (Since

$\cot x \times \tan x = 1, \cot 45^{\circ}$ )

$=1$

Hence, option

$A$ is correct.

$\cos\theta = \dfrac{14}{4}$ and

$\sin\theta$

$=$

$\dfrac{8}{3}$ , what is the value of

$\cot\theta$ ?

0%
$\dfrac{11}{16}$
0%
$\dfrac{13}{16}$
0%
$\dfrac{16}{21}$
0%
$\dfrac{21}{16}$

Explanation

$\cot \theta$

$=$

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

$=$

$\dfrac{\dfrac{14}{4}}{\dfrac{8}{3}}$

$=\dfrac{14}{4}\times\dfrac{3}{4}$

$\cot\theta = \dfrac{21}{16}$

So, option

$D$ is correct.

$\cos\theta = \dfrac{2}{21}$ and

$\sin\theta = \dfrac{6}{7}$ , what is the value of

$\tan\theta$ ?

0%
$4$
0%
$9$
0%
$6$
0%
$7$

Explanation

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

$=$

$\dfrac{\dfrac{6}{7}}{\dfrac{2}{21}}$

$=\dfrac{6}{7}\times\dfrac{21}{2}$

$\tan\theta = 9$

So, option

$B$ is correct.

$t = 45^{\circ}$ , what is

$\sec (t) \sin (t) - \mathrm{cosec} (t) \cos (t)$ ?

0%
$-2$
0%
$-1$
0%
$0$
0%
$\dfrac {\pi}{2}$
0%
None of the above

Explanation

For

$t=45^0$ , the value of

$\sec { (t) } \sin { (t) } -\csc { (t) } \cos { (t) }$ may be determined as follows:

$\sec { (t) } \sin { (t) } -\csc { (t) } \cos { (t) } =\sec { (45^{ 0 }) } \sin { (45^{ 0 }) } -\csc { (45^{ 0 }) } \cos { (45^{ 0 }) }$

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

$=1-1$

$=0$

$\sin \theta + \cos \theta = \sqrt {2}$ , find the value of

$\sin \theta \times \cos \theta$ .

0%
$2$
0%
$\sqrt {2} - 1$
0%
$\dfrac {1}{2}$
0%
$2\sec \theta$

Explanation

$\sin { \theta } +\cos { \theta } =\sqrt { 2 }$

Squaring both sides, we get

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

$\Rightarrow \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } .\cos { \theta } =2$

$\Rightarrow 1+2\sin { \theta } .\cos { \theta } =2$

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

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

Hence, the answer is

$\dfrac { 1 }{ 2 } .$

In the following figure

$AB \perp BC$ and

$\angle ACB = 30^\circ$ , given

$BC = \sqrt{300}\ m$ . The length of

$AB$ is

0%
$10\ m$
0%
$100\ m$
0%
$10 \sqrt 3\ m$
0%
$100 \sqrt 3\ m$

Explanation

Given that:

From the fig;

$AB\bot BC,\angle ACB=30^\circ, BC=\sqrt300 m$

To find:

$AB=?$

Solution:

$\triangle ABC,$

$\tan\angle ACB=\cfrac{AB}{BC}$

or,

$\tan30^\circ=\cfrac{AB}{BC}$

or,

$AB=\tan30^\circ\times BC$

or,

$AB=\cfrac{1}{\sqrt3}\times \sqrt{300}m$

or,

$AB=\cfrac{1}{\sqrt3}\times 10\sqrt{3}m$

or,

$AB=10m$

Therefore, A is the correct option.

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

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Practice Class 10 Maths Quiz Questions and Answers

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

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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