MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 2

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

$\theta$ in a right angled triangle.

0%
$\cos{\theta}$
0%
$\sin{\theta}$
0%
$\tan{\theta}$
0%
None of these

Explanation

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

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

$\text{sec}\ \theta = \dfrac{hypotenuse}{adjacent}$

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

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

The value of

$\displaystyle\frac{\cot{40^\circ}}{\tan{50^\circ}}-\frac{1}{2}\left(\frac{\cos{35^\circ}}{\sin{55^\circ}}\right)$ is ____________.

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

Explanation

$\displaystyle\frac{\cot{40^\circ}}{\tan{50^\circ}}-\frac{1}{2}\left(\frac{\cos{35^\circ}}{\sin{55^\circ}}\right)$

$=\displaystyle\frac{\cot{(90^\circ-50^\circ)}}{\tan{50^\circ}}-\frac{1}{2}\left(\frac{\cos{(90^\circ-55^\circ)}}{\sin{55^\circ}}\right)$

$=\displaystyle\frac{\tan{50^\circ}}{\tan{50^\circ}}-\frac{1}{2}\left(\frac{\sin{55^\circ}}{\sin{55^\circ}}\right)$

$=1-\dfrac 1 2$

$=\dfrac 1 2$

For an acute angle

$\theta$ in a right angled triangle,

$\dfrac{1}{\text{cosec}\,{\theta}} =$

0%
$\sin{\theta}$
0%
$\cos{\theta}$
0%
$\tan{\theta}$
0%
None of these

Explanation

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

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

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

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

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

The history of trigonometry goes back to the earliest recorded mathematics in Egypt and _____.

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German
0%
Indian
0%
Babylon
0%
Japanese

Trigonometry is used mainly due to the purpose of time keeping and _____.

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space
0%
stars
0%
astronomy
0%
planets

Trigonometry is a branch of mathematics that studies relationships involving lengths and ______ of triangles.

0%
radian
0%
degree
0%
angle
0%
vector

The term trigonometry was first invented by the German mathematician ______.

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

The first recorded use of trigonometry came from the Hellenistic mathematician ________________.

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

What is the value of

$\cos\theta \times\sec\theta$ ?

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

Explanation

$\cos\theta = \dfrac{1}{\sec\theta}$
Therefore,

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

Triangle measurement is called as _______.

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pythagoras
0%
trigonometry
0%
calculus
0%
area of square

Who is the founder of trigonometry?

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

______ mathematicians created the trigonometry system based on the sine function instead of the chords.

0%
Greek
0%
Indian
0%
German
0%
Egyptian

Explanation

Indian mathematicians

$\text{Aryabhata}$ created the trigonometry system based on the sine function instead of the chords.

Hence, the answer is Indian.

In the early 9th century AD, _________ produced accurate sine and cosine tables, and the first table of tangents.

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Habash al-Hasib al-Marwazi
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Muhammad ibn Jabir al-Harrani al-Battani
0%
Muhammad ibn Musa al-Khwarizmi
0%
Abu al-Wafa al-Buzjani

Given

$\tan \theta = 1,$ which of the following is not equal to

$\tan$

$\theta\ ?$

0%
$\sin 0^\circ$
0%
$\sin 90^\circ$
0%
$\cot 45^\circ$
0%
$\cos 0^\circ$

Explanation

Given:

$\tan \theta =1$

We observe that

$\sin \; 0^{\circ} =0$

$\sin \; 90^{\circ} =1$

$\cot \; 45^{\circ} =1$

$\cos \; 0^{\circ} =1$

Hence

$\sin \; 0^{\circ} \ne \tan \theta$

$\tan 5\theta . \tan4 \theta =1$ then

$\theta =$ ________.

0%
$10$
0%
$3$
0%
$7$
0%
$9$

Explanation

$\tan 5\theta . \tan4\theta =1$

$\therefore \tan 5\theta=\dfrac{1}{\tan4\theta}$

$\therefore \tan5 \theta = \cot4 \theta$

$\therefore\tan 5\theta =\tan (90^o-4\theta)$

$\therefore 5\theta =90^o-4\theta$

$\therefore 9\theta=90^o$

$\theta =10^o$

$\cos{{12}^{\circ}}-\sin{{78}^{\circ}}=$

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

Explanation

$\cos {12}^{\circ}= \sin{( 90 - 12)}^{\circ} = \sin{78}^{\circ}$

So,

$\cos(12^\circ) - \sin (78^\circ) = \sin(78^\circ) - \sin(78^\circ) = 0$

State the following statement is true or false

$cosec \theta = \sqrt {1 + \cot^{2}\theta}$

0%
True
0%
False

Explanation

We know that

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

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

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

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

Hence proved

The value of

$\sin 60^{\circ} - \cos 30^{\circ}$ is

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

Explanation

$\sin { 60° } -\cos { 30° }$

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

$=0$

Hence, the answer is

$0.$

Solve

$\dfrac{\tan{\theta}}{1-\cot{\theta}}+\dfrac{\cot{\theta}}{1-\tan{\theta}}$

0%
$1+\tan\theta+\cot\theta$
0%
$1-\tan\theta+\cot\theta$
0%
$1+\tan\theta-\cot\theta$
0%
None of these

Explanation

$\dfrac {{\tan \theta }}{{1 - \cot \theta }} + \dfrac {{\cot \theta }}{{1 - \tan \theta }}$

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

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

$= \dfrac {{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\sin \theta \cos \theta \left( {\sin \theta - \cos \theta } \right)}}$

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

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

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

$= 1 + \tan \theta + \cot \theta$

The given relation

$sin^2Acos^2B+cos^2Asin^2B+cos^2Acos^2B+sin^2Asin^2B=2$ is

0%
True
0%
False

Explanation

As we know that

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

$\sin^2 A\cos^2 B+\cos^2 A\sin^2 B+\cos^2 A\cos^2 B+\sin^2 A\sin^2 B=\sin^2 A(\cos^2 A+\sin^2 B)+\cos^2 A(\sin^2 B+\cos^2 B)$

$=\sin^2 A(1)+\cos^2 A(1)$

$=\sin^2 A+\cos^2 A$

$=1$

So the given relation is

$\text{False}$

$\dfrac{\sin 30^{o} \cos 60^{o}}{\sin 45^{o} \cos 45^{o}}=$

0%
$1$
0%
$\dfrac{1}{\sqrt{2}}$
0%
$\dfrac{1}{2}$
0%
$None\ of\ these$

Explanation

Consider the following question.

$\dfrac{\sin{{30}^{0}}\cos{{60}^{0}}}{\sin{{45}^{0}}\cos{{45}^{0}}}$

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

$=\dfrac{1}{2}$

Hence, this is the required answer.

$x=a\sin{\theta}$ and

$y=a\cos{\theta}$ then find the value of

${x}^{2}+{y}^{2}$

0%
$a^2$
0%
$4a^2$
0%
$2a^2$
0%
None of these

Explanation

$x^2+y^2=a^2sin^2\theta+a^2cos^2\theta\\=a^2(sin^2\theta+cos^2\theta)\\=a^2$

What is the value of

$\sin{45^{o}}+\cos{45^{o}}$

0%
$\sqrt {7}$
0%
$\sqrt {5}$
0%
$\sqrt {2}$
0%
$\sqrt {3}$

Explanation

According to question,

$sin45^o+cos45^o$

$\implies \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2}$

$\implies \dfrac{2}{\sqrt2}$

$\implies \sqrt2$

$sin{ 81 }^{ 0 }+tan{ 81 }^{ 0 }=.......$

0%
$cos{ 9 }^{ 0 }+cot{ 81 }^{ 0 }$
0%
$cos{ 9 }^{ 0 }+tan{ 81 }^{ 0 }$
0%
$cos{ 9 }^{ 0 }+cot{ 9 }^{ 0 }$
0%
$sin81^{ 0 }+cot{ 81 }^{ 0 }$

Explanation

$\sin 81^o + \tan 81^o$

$= \cos (90^o-81^o) + \cot (90^o - 81^o)$

$=\cos 9^o + \cot 9^o$

Hence, option C is correct.

From the given figure, AB =

0%
16
0%
8
0%
4
0%
12

Explanation

$\triangle ADC ,$

$sin30^{\circ} = \dfrac{DC}{AC}$

$\Rightarrow \dfrac{1}{2} = \dfrac{7x}{14}$

$\Rightarrow x = 1$

$BD = 4x = 4\times 1 = 4$

$\triangle ABD$

$sin30^{\circ} = \dfrac{BD}{AB}$

$\Rightarrow \dfrac{1}{2} = \dfrac{4}{AB}$

$\Rightarrow AB = 8$

${\rm{sin}}\,{\rm{\theta + cosec\theta = 2,}}\,{\rm{then}}\,\,{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta + cose}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\,$ is equal to:

0%
$1$
0%
$4$
0%
$2$
0%
None of these

Explanation

We have,

$\sin \theta +\csc \theta=2$

On squaring both sides, we get

$(\sin \theta +\csc \theta)^2=2^2$

$\sin^2 \theta +\csc^2 \theta+2\sin \theta\csc \theta=4$

$\sin^2 \theta +\csc^2 \theta+2\times 1=4$

$\sin^2 \theta +\csc^2 \theta+2=4$

$\sin^2 \theta +\csc^2 \theta=2$

Hence, this is the answer.

Solve :

$\dfrac { 2tan{ 30 }^{ \circ } }{ 1+{ tan }^{ 2 }{ 30 }^{ \circ } }$

0%
sin ${ 60 }^{ \circ }$
0%
cos ${ 60 }^{ \circ }$
0%
tan ${ 60 }^{ \circ }$
0%
cot ${ 60 }^{ \circ }$

Explanation

We know that,

$tan(2A) = \dfrac{2tanA}{1+tan^2A}$

Putting A to be

$30^o$

We get,

$tan(60^o) = \dfrac{2tan30^o}{1+tan^230^o}$

Thus, C is the correct answer.

$\sin x=\dfrac 12\\\tan x=?$

0%
$\dfrac{1}{\sqrt{3}}$
0%
$\dfrac{1}{\sqrt{2}}$
0%
$\dfrac{1}{\sqrt{6}}$
0%
$\dfrac{1}{\sqrt{5}}$

Explanation

$\sin x=\dfrac 12\\\sin x=\sin 30^{\circ}\\x=30\\\tan 30=\dfrac 1 {\sqrt3}$

sin0

$^0$ . sin1

$^0$ ..................... sin90

$^0$ =

0%
0
0%
tan0 $^0$
0%
csc90 $^0$
0%
cos90 $^0$

Explanation

Given,

$\sin 0 \cdot \sin 1.....................\cdot \sin 90$

Since,

$\sin 0=0$

Hence the product is

$0$

$\displaystyle \cot\theta=3x-\frac{1}{12{x}}$ , then

$\csc \theta-\cot \theta$ is equal to

0%
$6x$
0%
$-\displaystyle \frac{1}{6x}$
0%
$\displaystyle \frac{1}{6x}$ or $-6x$
0%
$6x$ or $-\displaystyle \frac{1}{6x}$

Explanation

Given:

$\cot \theta=3x-\dfrac {1}{12x}$

We know

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

$\Rightarrow \csc^2 \theta= 1+\cot^2 \theta$

$\Rightarrow \csc^2 \theta=1+\bigg( 3x-\dfrac{1}{12x} \bigg)^2$

$\Rightarrow \csc^2 \theta=1+9x^2+\dfrac{1}{144x^2}-\dfrac {1}{2}$

$\Rightarrow \csc^2 \theta=9x^2+\dfrac{1}{144x^2}+\dfrac {1}{2}$

$\Rightarrow \csc^2 \theta=\bigg( 3x+\dfrac{1}{12x} \bigg)^2$

$\Rightarrow \csc \theta=\pm \bigg( 3x+\dfrac{1}{12x} \bigg)$

$\csc \theta=3x+\dfrac{1}{12x}$

Then

$\csc \theta - \cot \theta=3x+\dfrac{1}{12x}-\bigg(3x-\dfrac{1}{12x}\bigg)$

$=\dfrac{1}{6x}$

$\csc \theta=-\bigg(3x+\dfrac{1}{12x}\bigg)=-3x-\dfrac{1}{12x}$

Then

$\csc \theta - \cot \theta=-3x-\dfrac{1}{12x}-\bigg(3x-\dfrac{1}{12x}\bigg)$

$=-6x$

Hence

$\csc \theta-\cot \theta=-6x$ or

$\dfrac{1}{6x}$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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