MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 1

The value of

$\displaystyle\,\frac{2}{tan\,30^{\circ}}$ is ________

0%
3.46
0%
3.56
0%
34.6
0%
3.48

Explanation

$\dfrac{2}{\tan30^{\circ}} = \tfrac{2}{\frac{1}{\sqrt{3}}} = 2 \sqrt{3} = 3.46$

The value of

$\tan\theta \times \cot\theta =$ _____

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

Explanation

To find: Value of

$\tan\theta \times \cot\theta$
We know that

$\cot\theta = \dfrac{1}{\tan\theta}$
So,

$\tan\theta \times \dfrac{1}{\tan\theta} = 1$

Which of the following is equal to

$\sin x \sec x$ ?

0%
$\tan x$
0%
$\cot x$
0%
$\cos x \tan x$
0%
$\cos x \text{cosec}\, x$
0%
$\cot x \text{cosec}\, x$

Explanation

$\sin { x }\cdot \sec { x }$

$=\sin { x } \times\dfrac { 1 }{ \cos x }$

$=\dfrac { \sin { x } }{ \cos x }$

$=\tan { x }$

The value of

$\sin\theta\times \text{cosec}\,\theta =$ _____

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

Explanation

$\sin\theta \times cosec\;\theta$

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

$=1$

Hence, the answer is

$1.$

Who published the trigonometry in 1595?

0%
William Rowan Hamilton
0%
Hipparchus
0%
Bartholomaeus Pitiscus
0%
Newton

Choose the correct alternative answer for the following question.

$cosec \ 45^\circ= ?$

0%
$\dfrac12$
0%
$\sqrt{2}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
$\dfrac{2}{\sqrt{3}}$

Explanation

$\sin 45^{\circ}=\dfrac {1}{\sqrt2}$

$cosec \ 45^\circ = \sqrt{2}$

Hence, the correct answer is

$\sqrt{2}$ .

$\sqrt{3} \cos A = \sin A$ , then the value of

$\cot A$ is:

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

$\tan {45^ \circ } =?$

0%
$1$
0%
$\sqrt3$
0%
$0$
0%
can't be determined

Explanation

$\tan 45^{\circ}=\dfrac{\sin 45^{\circ}}{\cos 45^{\circ}}$

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

$=1$

So the relation is

$\text{True}$

The value of

$\cos 1^o.\cos 2^o.\cos 3^o.....\cos 179^o$ is equal to

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

Explanation

$A = \cos1^0\cdot\cos2^0\cdot\cos3^0\cdots\cos{90}^0\cdots\cos{179}^0$

$\because \cos{90}^0 = 0$

$\therefore A = 0$

Hence, option B.

$\triangle ABC$ ,

$\angle B= {90}^{\circ}$ ,

$\sin{C} = \dfrac{3}{5}$ , then

$\cos{A} =$ ______

0%
$\dfrac{3}{5}$
0%
$\dfrac{4}{5}$
0%
$\dfrac{5}{4}$
0%
$\dfrac{5}{3}$

Explanation

In a right angled triangle,

$A+B+C=180^\circ$

$B=90^\circ$

So,

$A+C=90^\circ$

$A=90^\circ-C$

$\cos A=\cos(90^\circ-A)$

$\Rightarrow \cos A=\sin C=\dfrac35$

$\Rightarrow \cos A=\cfrac35$

If sin

$A=\dfrac{1}{2}$ and cos

$B=\dfrac{1}{2}$ , then the value of

$(A+B)$ is equal to :

0%
$0^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$
0%
$30^{\circ}$

Explanation

$\sin { A } =\dfrac { 1 }{ 2 } ,\cos { B } =\dfrac { 1 }{ 2 }$

$\Rightarrow \sin { A } =\sin { 30° } ,\cos { B } =\cos { 60° }$

$\Rightarrow A=30°,B=60°$

$\therefore A+B=30°+60°=90°$

Hence, the answer is

$90°.$

If cot

$\theta =\dfrac{7}{8}$ , then the value of

$tan^{2}\theta$ equals to :

0%
$\dfrac{8}{7}$
0%
$\dfrac{49}{64}$
0%
$\dfrac{64}{49}$
0%
$\dfrac{7}{8}$

Explanation

$\cot { \theta } =\dfrac { 7 }{ 8 }$

$\Rightarrow \tan { \theta } =\dfrac { 1 }{ \cot { \theta } } \\\ \ \ \ \ \ \ \ \ \ \ \ =\dfrac { 8 }{ 7 }$

$\Rightarrow \tan ^{ 2 }{ \theta } =\dfrac { 64 }{ 49 }$

Hence, the correct option is (C)

The value of (sin

$45^{\circ}+cos 45^{\circ}$ ) is :

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

Explanation

$\sin { 45° } +\cos { 45° }$

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

$=\dfrac { 2 }{ \sqrt { 2 } }$

$=\sqrt { 2 }$

Hence, the answer is

$=\sqrt { 2 } .$

$2 \cos\theta+ \sin\theta=1$ , then the value of

$4 \cos\theta + 3 \sin\theta$ is equal to

0%
$3$
0%
$-5$
0%
$\dfrac {7}{5}$
0%
$-4$

$\cos ^{2} 60^{\circ}+\sin ^{2} 30^{\circ} = \displaystyle \frac{1}{a}$
Then

$a=$

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

Explanation

$\cos^2 60^{\circ} + \sin^2 30^{\circ}$

$=\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2$

$=\cfrac{1}{4} + \cfrac{1}{4}$

$=\cfrac{2}{4}$

$=\cfrac{1}{2}$

$\therefore \dfrac1a = \dfrac12$

$\Rightarrow a = 2$

$\sec 4A = cosec (A-20^{\small\circ})$ , where

$4A$ is an acute angle, find the value of

$A$ .

0%
$A = 32^{\small\circ}$
0%
$A = 22^{\small\circ}$
0%
$A = 41^{\small\circ}$
0%
$A = 16^{\small\circ}$

Explanation

Hence

$\sin(A-20^{0})=\cos(4A)$

$\sin(A-20^{0})=\sin(90^{0}-4A)$
Hence

$A-20^{0}=90^{0}-4A$

$5A=110^{0}$

$A=22^{0}$

In the given figure,

$BC=15cm$ and

$\sin{B}=\dfrac{4}{5}$ , What is the value of

$AB$ ?

0%
$25cm$
0%
$20cm$
0%
$5cm$
0%
$4cm$

Explanation

$\sin B= \dfrac{AC}{AB} =\dfrac{\sqrt{AB^2 - BC^2}}{AB}$

$\dfrac{4}{5} AB= \sqrt{AB^2 - 15^2}$

$\Rightarrow 16AB^2 = 25AB^2 - (25)(15)^2$

$AB^2 = \dfrac{25\times 15 \times 15}{9}$

$\Rightarrow \boxed {AB = 25}$

Choose the correct option. Justify your choice.

$\displaystyle 9{ \sec }^{ 2 }A-9{ \tan }^{ 2 }A=$

0%
$1$
0%
$9$
0%
$8$
0%
$0$

Explanation

We know,

$1= \sec ^2A-\tan^2A$

$\therefore 9(\sec^2A-\tan^2A)$

$=9\times 1=9$

$sin({ 90 }^{ 0 }-\theta )=\dfrac { 3 }{ 7 }$ , then

$cos\theta$

0%
$\cfrac { 7 }{ 3 }$
0%
$\cfrac { 3 }{ 7 }$
0%
$\cfrac { -7 }{ 3 }$
0%
$\cfrac { -3 }{ 7 }$

Explanation

$\sin (90^o - \theta) = \dfrac{3}{7}$

We know,

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

Hence,

$\cos \theta = \dfrac{3}{7}$

$\tan \theta= \cot \theta$ , then the value of

$\sec \theta$ is :

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

Explanation

$\tan { \theta } =\cot { \theta }$

$\Rightarrow \tan ^{ 2 }{ \theta } =1$

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

$\Rightarrow \theta =45°$

$\therefore \sec { 45° } =\sqrt { 2 }$

Hence, the answer is

$\sqrt { 2 }.$

$\sqrt{3}\,=\,1.732$ , find (correct to two decimal places) the value of each of the following :

$sin\,60^{\circ}$ is 0.87
State true or false.

0%
True
0%
False

Explanation

$\sin 60^{\circ} = \dfrac{\sqrt{3}}{2} = \dfrac{1.732}{2} = 0.87$

The value of cosec

$30^{\circ}$ + cot

$45^{\circ}$ is:

0%
2
0%
1
0%
3
0%
-1

Explanation

cosec

$30^{\circ}$ + cot

$45^{\circ}$

$=2+1$

$=3$

Hence, the answer is

$3.$

The value of

$5cosec^{2}\theta -5cot^{2 }\theta$ is equal to:

0%
5
0%
1
0%
0
0%
-5

Explanation

Using the identity:

$cosec^{2}\theta -\cot^{2 }\theta=1$

We get,

$5cosec^{2}\theta -5\cot^{2 }\theta$ = 5(

$cosec^{2}\theta -\cot^{2 }\theta)=5$

$\theta =45^{\circ},$ then the value of

$cosec^{2}\theta$ is

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

Explanation

$\theta =45°$

$\Rightarrow cosec^{ 2 }45°={ \left( \sqrt { 2 } \right) }^{ 2 }=2$

Hence, the answer is

$2.$

For an acute angle

$\theta$ in a right angled triangle,

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

0%
$\cos{\theta}$
0%
$\cot{\theta}$
0%
$\sec{\theta}$
0%
$\text{cosec}\,{\theta}$

Explanation

$\sin \theta=\dfrac{opposite}{hypotenuse}$

$\cos \theta = \dfrac{adjacent}{hypotenuse}$

$\dfrac{1}{\sin \theta}=\dfrac{1}{\dfrac{opposite}{hypotenuse}}=\dfrac{hypotenuse}{opposite}=\text{cosec}\ \theta$

$\therefore\ \dfrac{1}{\sin \theta}=\text{cosec}\ \theta$

What is the value of

$\displaystyle \sin { { 30 }^{ o } } \times co\sec{ 30 }^{ o }$ ?

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

Explanation

$\displaystyle \sin { { 30 }^{ o }\times co\sec{ 30 }^{ o } } =\cfrac { 1 }{ 2 } \times \cfrac { 2 }{ 1 } =\cfrac { 1 }{ 1 } =1$

The value of

$\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }$ is :

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

Explanation

$\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }=1\times 1=1$

$\dfrac{1}{\cos{\theta}} =$ _____ , for an acute angle

$\theta$ in a right angled triangle.

0%
$\text{cosec}\,{\theta}$
0%
$\sec{\theta}$
0%
$\cot{\theta}$
0%
None of these

Explanation

$\sin \theta=\dfrac{opposite}{hypotenuse}$

$\cos \theta = \dfrac{adjacent}{hypotenuse}$

$\dfrac{1}{\cos \theta}=\dfrac{1}{\dfrac{adjacent}{hypotenuse}}=\dfrac{hypotenuse}{adjacent}=\sec \theta$

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

The value of

$\tan{\theta}\cdot\tan{(90-\theta)}$ is equal to

0%
$\sin^2{\theta}$
0%
$1$
0%
$\cos^2{\theta}$
0%
$0$

Explanation

$\tan\theta \cdot \tan(90-\theta) = \tan\theta \times \cot\theta = 1$

$\because \tan(90-\theta) =\cot \theta$

$\sec{2A}=\csc{(A-42^\circ)}$ where

$2A$ is acute angle then value of

$A$ is

0%
$44^\circ$
0%
$22^\circ$
0%
$21^\circ$
0%
$66^\circ$

Explanation

$\sec{2A}=\csc{(A-42^\circ)}$

$\csc{(90-2A)}=\csc{(A-42^\circ)}$

$90-2A=A-42$

$3A=132$

$A=44^\circ$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 1 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 1 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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