MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 4

The value of

$\sin^2 60^0 + \tan 45^0 - \text{cosec}^2 45^0$ is

0%
$-\dfrac{1}{2}$
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$-\dfrac{1}{4}$
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$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$

Explanation

$\sin^2 60^0 + \tan 45^0 - \text{cosec}^2 45^0$

$=\left(\dfrac{\sqrt 3}{2}\right)^2+1-(\sqrt 2)^2=\dfrac34-1=\dfrac{-1}{4}$

$\dfrac{tan \theta}{(1+tan^{2} \theta)^2}+\dfrac{cot \theta}{(1+cot^{2} \theta)^2}$ is equal to

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$2 sin \theta. cos \theta$
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$cosec \theta. sec \theta$
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$sin \theta. cos \theta$
0%
$2 cosec \theta. sin \theta$

Explanation

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

$[\therefore sec^2 \theta =1 +tan^2 \theta, cosec^2 \theta=1+cot^2 \theta]$

$= sin \theta cos^3 \theta + cos \theta sin^3 \theta$

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

$= sin \theta cos \theta$

Let

$x = (1 + \sin {A})(1 + \sin {B})(1 + \sin {C}), y = (1 - \sin {A})(1 - \sin {B})(1 - \sin {C})$ and If

$x = y$ then

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$x = \cos {A} \cos {B} \cos {C}$
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$y = \sin {A} \sin {B} \sin {C}$
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$y = -\cos {A} \cos {B} \cos {C}$
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$x = - \sin {A} \sin {B} \sin {C}$

Explanation

Given :

$x = (1 + \sin {A})(1 + \sin {B})(1 + \sin {C})$ .......

$(i)$ and

$y = (1 - \sin {A})(1 - \sin {B})(1 - \sin {C})$ .......

$(ii)$

On multiplying both the equation we get

$xy = (1 - \sin^{2} A)(1 - \sin^{2} B)(1 - \sin^{2} C)$

$\therefore xy = \cos^{2} A \cos^{2} B \cos^{2} C$

$x = y,$ then

$x^{2} = \cos^{2} A \cos^{2} B \cos^{2} C$

$\implies x = \pm \cos {A} \cos {B} \cos {C}$

$\implies y = \pm \cos {A} \cos {B} \cos {C}$

Hence, option A and C are correct.

$a \cos\theta-b \sin\theta=c$ , then

$a \sin\theta+b\cos\theta$ equals-

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$a^2+b^2+c^2$
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$a^2-b^2-c^2$
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$\sqrt {a^2+b^2+c^2}$
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$\pm \sqrt {a^2+b^2-c^2}$

Explanation

$a\cos\theta - b\sin\theta = c$

Let,

$a\sin\theta + b\cos\theta = x$

Squaring and adding we get

$a^2\cos^2 \theta+b^2 \sin^2 \theta-2ab\sin \theta \cos \theta+a^2 \sin^2 \theta+b^2 \cos^2 \theta+2ab\sin \theta \cos \theta=c^2+x^2$

$\Rightarrow a^2 + b^2 -c^2 = x^2$

$x=\pm \sqrt{a^2+b^2-c^2}$

Option D is the correct option

$\displaystyle \frac { \sin ^{ 4 }{ x } }{ 2 } +\frac { \cos ^{ 4 }{ x } }{ 3 } =\frac { 1 }{ 5 } ,$ then:

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$\displaystyle \tan ^{ 2 }{ x } =\frac { 2 }{ 3 }$
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$\displaystyle \frac { \sin ^{ 8 }{ x } }{ 8 } +\frac { \cos ^{ 8 }{ x } }{ 27 } =\frac { 1 }{ 125 }$
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$\displaystyle \tan ^{ 2 }{ x } =\frac { 1 }{ 3 }$
0%
$\displaystyle \frac { \sin ^{ 8 }{ x } }{ 8 } +\frac { \cos ^{ 8 }{ x } }{ 27 } =\frac { 2 }{ 125 }$

Explanation

$\textbf{Step 1: Simplify the given equation using basic trigonometric identities}$

$\text{We have the given expression as}$

$\dfrac{\sin^4 {x}}{2} + \dfrac{\cos^4{x}}{3}=\dfrac{1}{5}$

$\Rightarrow \dfrac{\sin^4 {x}}{2} + \dfrac{\cos^4{x}}{3}=\dfrac{1}{5}$

$\Rightarrow \dfrac{\sin^4 {x}}{2} + \dfrac{(1-\sin^2{x})^2}{3}=\dfrac{1}{5}$

$\boldsymbol{[\because \sin^2 x + \cos^2 x =1]}$

$\Rightarrow \dfrac{\sin^4 {x}}{2} + \dfrac{(\sin^4{x} +1-2\sin^2{x})}{3}=\dfrac{1}{5}$

$\Rightarrow 25\sin^4{x}-20\sin^2{x}+4=0$

$\Rightarrow (5\sin^2{x}-2)^2=0$

$\Rightarrow \sin^2{x}=\dfrac{2}{5}$

$\Rightarrow\cos^2{x}= \dfrac{3}{5}$

$\Rightarrow \tan^2{x}= \dfrac{2}{3}$

$\boldsymbol{[\because \tan^2 x =\dfrac{\sin^2 x}{\cos^2 x}]}$

$\textbf{Step 2: Find the value of}$

$\boldsymbol{\displaystyle \frac { \sin ^{ 8 }{ x } }{ 8 } +\frac { \cos ^{ 8 }{ x } }{ 27 }}$

$\text{Also putting the newfound values of sin x and cos x in the eq. we get}$

$\Rightarrow \dfrac{\sin^8{x}}{8} + \dfrac{\cos^8{x}}{27}= \dfrac{(\sin^2{x})^4}{8} + \dfrac{(\cos^2{x})^4}{27}$

$\Rightarrow \dfrac{(\dfrac{2}{5})^4}{8} +\dfrac{(\dfrac{3}{5})^4}{27}=$

$\displaystyle\dfrac{16}{625\times 8}+\dfrac{81}{625\times27}=\dfrac{1}{625}\times\dfrac{27\times16+81\times8}{8\times27}=\frac5{625}=\dfrac { 1 }{ 125 } .$

$\textbf{Hence option A and B are correct.}$

$\displaystyle x=r\sin \theta \cdot \cos \phi,$

$y=r\sin \theta \cdot \sin \phi$ and

$\displaystyle z= r\cos \theta$ , then the value of

$\displaystyle x^{2}+y^{2}+z^{2}$ is independent of:

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$\displaystyle \theta ,\phi$
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$\displaystyle r,\theta$
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$\displaystyle r,\phi$
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$r$

Explanation

Given,

$\displaystyle x=r\sin \theta \cdot \cos \phi,$

$y=r\sin \theta \cdot \sin \phi$ and

$\displaystyle z= r\cos \theta$

Therefore,

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

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

$={ r }^{ 2 }{ \sin }^{ 2 }\theta { \cos }^{ 2 }\phi +{ r }^{ 2 }{ \sin }^{ 2 }\theta { \sin }^{ 2 }\phi +{ r }^{ 2 }{ \cos }^{ 2 }\theta$

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

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

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

$={ r }^{ 2 }$

$\sin { A } =a\cos { B }$ and

$\cos { A } =b\sin { B }$ , then

$\left( { a }^{ 2 }-1 \right) \tan ^{ 2 }{ A } +\left( 1-{ b }^{ 2 } \right) \tan ^{ 2 }{ B }$ is equal to

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$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 } }{ { b }^{ 2 } }$
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$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }$
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$\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } }$
0%
$\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 } }$

Explanation

Given,

$\sin { A } =a\cos { B }$ and

$\cos { A } =b\sin { B } ,$

On squaring and adding, we get

${ a }^{ 2 }\cos ^{ 2 }{ B } +{ b }^{ 2 }\sin ^{ 2 }{ B } =1$

$\Rightarrow \displaystyle \cos ^{ 2 }{ B } =\frac { 1-{ b }^{ 2 } }{ { a }^{ 2 }-{ b }^{ 2 } }$ and

$\displaystyle \sin ^{ 2 }{ B } =\frac { { a }^{ 2 }-1 }{ { a }^{ 2 }-{ b }^{ 2 } } ,$

$\Rightarrow \displaystyle \tan ^{ 2 }{ B } =\frac { { a }^{ 2 }-1 }{ 1-{ b }^{ 2 } }$

Also,

$\displaystyle\tan ^{ 2 }{ A } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \cot ^{ 2 }{ B } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } .\frac { 1-{ b }^{ 2 } }{ { a }^{ 2 }-1 }$

Therefore,

$\displaystyle\left( { a }^{ 2 }-1 \right) \tan ^{ 2 }{ A } +\left( 1-{ b }^{ 2 } \right) \tan ^{ 2 }{ B } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( 1-{ b }^{ 2 } \right) +{ a }^{ 2 }-1=\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { b }^{ 2 } }$

$\displaystyle \sin\theta + cosec \theta= 2,$ then the value of

$\displaystyle \sin ^{8}\theta + cosec ^{8}\theta$ is equal to

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$2$
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$\displaystyle 2^{8}$
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$\displaystyle 2^{4}$
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none of these

Explanation

$sin\theta +cosec\theta =2\\ \Rightarrow sin\theta +\cfrac { 1 }{ sin\theta } =2$
Squaring both sides

${ \left( sin\theta +\cfrac { 1 }{ sin\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { sin }^{ 2 }\theta +\cfrac { 1 }{ { sin }^{ 2 }\theta } +2sin\theta \cfrac { 1 }{ sin\theta } =4\\ \Rightarrow { sin }^{ 2 }\theta +\cfrac { 1 }{ { sin }^{ 2 }\theta } =2$
Again squaring both sides

${ \left( { sin }^{ 2 }\theta +\cfrac { 1 }{ { sin }^{ 2 }\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { sin }^{ 4 }\theta +\cfrac { 1 }{ { sin }^{ 4 }\theta } =2$
Similarly,

${ sin }^{ 8 }\theta +\cfrac { 1 }{ { sin }^{ 8 }\theta } =2$

$\Rightarrow { sin }^{ 8 }\theta +{ cosec }^{ 8 }\theta =2$

$\tan A+\cot A=4$ , then

$\tan^2A+\cot^2A$ is equal to

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$10$
0%
$11$
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$12$
0%
$14$

Explanation

Given,

$\tan A+\cot A=4$

On squaring, we get

$(\tan A+\cot A)^2=4^2$

$\Rightarrow \tan^2A+\cot^2A+2 \tan A \cot A=16$

$\Rightarrow \tan^2A+\cot^2A+2 (1) = 16$

$\Rightarrow \tan^2A+\cot^2A+2 =16$

$\Rightarrow \tan^2A+\cot^2A =14$

$\sin A, \cos A$ and

$\tan A$ are in

$G.P.,$ then

$\cot^6 A- \cot^2A$ is equal to

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

Explanation

Given,

$\sin A, \cos A$ and

$\tan A$ are in

$G.P.,$

$\therefore \cos^2A=\sin A \tan A$

$\Rightarrow \cos^2A=\sin A\times \dfrac{\sin A}{\cos A}$

$\Rightarrow \cos^3 A=\sin^2 A$

$\Rightarrow \cos^2A . \cos A = \sin^2 A$

$\Rightarrow \cfrac {\cos^2A}{\sin^2A} = \cfrac {1}{\cos A}$

$\Rightarrow \cot^2 A= \sec A$

Squaring on both sides, we get

$\Rightarrow (\cot^2 A)^2= (\sec A)^2$

$\Rightarrow \cot^4 A= \sec^2 A$

$\Rightarrow \cot^4A=1+\tan^2A$

$\Rightarrow \cot^4A-1 = \tan^2A......(1)$

Now,

$\cot^6A - \cot^2A$

$= \cot^2A(\cot^4A-1)$

$=\cot^2A \times \tan^2A$

$=\dfrac{1}{\tan^2A} \times \tan^2A$

$=1$

Hence, the required value is

$1.$

The value of

$\tan 5^{\circ}\tan 10^{\circ}\tan 15^{\circ}\cdots \tan 85^{\circ}$ is

0%
$0$
0%
Not defined
0%
$1$
0%
$-1$

Explanation

Using,

$\tan x=\cot \left( 90^{\circ}-x \right)$ and $\tan x.\cot x=1$

We get ,

$\tan 5.\tan 10.\tan 15...\tan 80.\tan 85\\ =\tan 5.\tan 10.\tan 15...\tan 45. \cot(90-35)...\cot \left( 90-10 \right) .\cot \left( 90-5 \right) \\ =\tan 5.\cot 5.\tan 10.\cot 10...\tan 45\\ =1$

The expression

$\displaystyle f\left ( \theta \right )= \frac{1}{\tan \theta+\sec \theta+\cot \theta+\text{cosec} \theta}$ is equivalent to

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$\displaystyle \frac{\text{cosec} \theta +\sec \theta+\text{cosec} \theta\sec \theta}{2\text{cosec} \theta\sec\theta}$
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$\displaystyle \frac{\text{cosec} \theta -\sec \theta-\text{cosec} \theta\sec \theta}{2\text{cosec}\theta\sec\theta}$
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$\displaystyle \frac{\text{cosec} \theta +\sec \theta-\text{cosec} \theta\sec \theta}{2\text{cosec}\theta\sec\theta}$
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None of these

Explanation

$\displaystyle f(\theta ) = \frac{1}{\tan \theta + \sec \theta + \cot \theta + \text{cosec } \theta }$

$\displaystyle= \frac{1}{\frac{\sin \theta }{\cos \theta} +\frac{ 1}{\cos \theta} + \frac{\cos \theta }{\sin \theta} + \frac{1}{\sin \theta} }$

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

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

$\displaystyle= \frac{\cos \theta + \sin \theta - 1}{2} = \frac{\text{cosec } \theta + \sec \theta - \text{cosec } \theta .\sec \theta }{2\text{cosec } \theta .\sec \theta }$ .....(.dividing by

$\cos \theta\sin\theta$ )

$ABC$ is an isosceles right-angled triangle. Assuming

$AB= BC= x$ , find the value of each of the following trigonometric ratios:

$\sin 45^{0}$ is

$\displaystyle \dfrac{1}{\sqrt{m}}$ , the value of

$m$ is

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

Explanation

Given:

$\triangle ABC$ ,

$AB = BC = x$ ,

$\angle ABC = 90^{\circ}$

By Pythagoras Theorem,

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

$AC^2 = x^2 + x^2$

$AC^2 = 2x^2$

$AC = \sqrt{2} x$

Since,

$AB = BC$ , then by Isosceles triangle property,

$\angle ACB = \angle CAB = 45^{\circ}$

Now,

$Sin 45^{\circ} = Sin C = \dfrac{P}{H}$

$= \dfrac{AB}{AC}$

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

so, the value of

$m$ is

$=2$

Choose the correct option for the following.

The value of

$\displaystyle \frac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$ is

$\displaystyle \frac{1-\sqrt{2}}{1+\sqrt{6}}$ , when

$A= 15^{0}$ .

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Given,

$A = 15^{\circ}$

Thus,

$\displaystyle \dfrac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$

Put

$A = 15^{\circ}$

$\displaystyle \dfrac{\cos 45^0 -2\cos 60^0}{\sin 45^0 + 2\sin 60^0}$

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

$\displaystyle \dfrac{\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}} + \sqrt{3}}$

$\displaystyle \dfrac{\dfrac{1 - \sqrt{2}}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}} + \sqrt{3}}$

$\displaystyle \dfrac{1 - \sqrt{2}}{1 + \sqrt{6}}$

$\tan \angle ADC= 1$ and

$\sin \angle ABC=4:5$ , then measure of

$CD$ is:

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

In $\triangle ABC$ , $BC = 15$ , $\sin B = \dfrac {4}{5}$

Now, $\sin B = \dfrac {P}{H}$
$\dfrac {4}{5} = \dfrac {AC}{AB}$
Then, let $AC = 4x$ and $AB = 5x$

Now, $AB^2 = AC^2 + BC^2$
$(5x)^2 = (4x)^2 + 15^2$
$25x^2 = 16x^2 + 225$
$9x^2 = 225$
$x = 5$

Hence, $AC= 20$ and $AB = 25$

$\tan \angle ADC = 1$
$\dfrac {AC}{CD} = 1$
$AC = CD = 20$

Now, using Pythagoras theorem,
$AD^2 = AC^2 + CD^2$
$AD^2 = 20^2 + 20^2$
$AD = 20 \sqrt{2}$

Find the length of

$AB$ in cm

0%
39.25
0%
81.96
0%
46.32
0%
85.34

Explanation

Given,

$BC = 30$

$\angle AED = \angle ABC = 90$
Thus,

$DE = BC = 30$

Now, In

$\triangle ADE$

$\tan 45 = \frac{AE}{DE}$

$1 = \frac{AE}{30}$

$AE = 30$

Now, In

$\triangle DBE$

$\tan 60 = \frac{BE}{DE}$

$BE = 30 \sqrt{3}$

Now,

$AB = BE + AE$

$AB = 30\sqrt{3} + 30$

$AB = 81.96$ cm

The length of

$AD$ (in cm.) is

0%
$10$
0%
$100$
0%
$75$
0%
$50$

Explanation

$\triangle ABC$

$BC = AC$
Thus,

$\angle ABC = \angle ACB = x$ (Isosceles triangle Property)

Now,

$\angle ACD = \angle ABC + \angle ACB$ (Exterior angle property)

$60 = x + x$

$x = 30^{\circ}$

Now,

$\sin \angle ABC = \cfrac{P}{H}$

$\sin 30 = \cfrac{AD}{AB}$

$\cfrac{1}{2} = \cfrac{AD}{100}$

$AD = 50$ cm

In the given figure;

$BC= 15$ cm and

$\displaystyle \sin B= \frac{4}{5}$ .
The measure of

$AC$ in cm is

0%
$10$
0%
$15$
0%
$20$
0%
$19$

Explanation

Find

$PQ$ , if

$\displaystyle AB = 150\ m, \angle P = 30^{\circ}$ and

$\displaystyle \angle Q = 45^{\circ}$

0%
$978.8$
0%
$524.4$
0%
$279.9$
0%
$409.8$

Explanation

Given,

$AB = 30$ ,

$\angle P = 30^{\circ}$ ,

$\angle Q= 45^{\circ}$

$\triangle APB$ ,

$\tan P = \cfrac{P}{B}$

$\tan 30 = \cfrac{AB}{PB}$

$\cfrac{1}{\sqrt{3}} = \cfrac{150}{PB}$

$PB = 150 \sqrt{3}$

$\triangle ABQ$ ,

$\tan Q = \cfrac{P}{B}$

$\tan 45 = \cfrac{AB}{BQ}$

$1 = \cfrac{150}{BQ}$

$BQ = 150$

Thus,

$AB = PB +BQ$

$AB = 150\sqrt{3} + 150$

$AB = 409.8$

If in given figure

$AD= 2$ cm, then value of

$AB$ ( in cm ) is

0%
2
0%
3
0%
7
0%
1

Explanation

Given,

$AD \perp CE$ ,

$\angle ACD = 45^{\circ}$ ,

$AD =2$

Now, In

$\triangle ADC$

$\tan \angle ACD = \dfrac{P}{B} = \dfrac{AD}{CD}$

$\tan 45 = \dfrac{2}{CD}$

$CD = 2=AB$

$AB=2$ cm

In the given figure,

$\displaystyle \angle B = 60^{\circ}, \angle C =30^{\circ}, AB=8\ cm$ and

$\displaystyle BC=25\ cm$ .

Calculate

$BE$ in cm.

0%
4
0%
2
0%
5
0%
7

Explanation

$\angle B = 60$ ,

$\angle C = 30$ ,

$AB = 8$ ,

$BC = 25$

Now, In

$\triangle ABE$

$\cos B = \frac{B}{H} = \frac{BE}{AB}$

$\cos 45 = \frac{BE}{8}$

$BE = 4$

Find :

$AD$

0%
2
0%
6
0%
4
0%
1

Explanation

Given:

$AB = 12$ ,

$\angle ACB = 30^{\circ}$

$\angle ACB + \angle ABC + \angle BAC = 180$

$30 + 90 + \angle BAC = 180$

$\angle BAC = 60^{\circ}$

Now, In

$\triangle ABD$
Now,

$\cos 30 = \frac{B}{H} = \frac{AD}{AB}$

$\frac{1}{2} = \frac{AD}{12}$

$AD = 6$

$\sin x+\sin^2x+\sin^3x=1$ , then

$\cos^6x-4\cos^4x+8 \cos^2x$ is equal to

0%
$0$
0%
$2$
0%
$4$
0%
$8$

Explanation

$\sin x+{ \sin }^{ 2 }x+{ \sin }^{ 3 }x=1\Rightarrow \sin x+{ \sin }^{ 3 }x=1-{ \sin }^{ 2 }x\\ \Rightarrow \sin x\left( 1+\sin ^{ 2 }x \right) ={ \cos }^{ 2 }x\Rightarrow \sin x\left( 2-{ \cos }^{ 2 }x \right) ={ \cos }^{ 2 }x\\ \Rightarrow { \sin }^{ 2 }x{ \left( 2-{ \cos }^{ 2 }x \right) }^{ 2 }={ \cos }^{ 4 }x\\ \Rightarrow \left( 1-{ \cos }^{ 2 }x \right) \left( 4+{ \cos }^{ 4 }x-4{ \cos }^{ 2 }x \right) ={ \cos }^{ 4 }x\\ \Rightarrow 4+{ \cos }^{ 4 }x-4{ \cos }^{ 2 }x-4{ \cos }^{ 2 }x-{ \cos }^{ 6 }x+4{ \cos }^{ 4 }x={ \cos }^{ 4 }x\\ \Rightarrow { \cos }^{ 6 }x-4{ \cos }^{ 4 }x+8{ \cos }^{ 2 }x=4$

$\cos 1^o \cos 2^o \cos 3^o ... \cos {179}^o=x+1$ , then

$x$ is equal to

0%
$-1$
0%
$0$
0%
$1$
0%
none of these

Explanation

$\cos1^{0}.\cos2^{0}...\cos90^{0}...\cos(179^{0})$

$=0$ since

$\cos90^{0}=0$
Hence

$x+1=0$

$x=-1$

For all values of

$\alpha$ the given expression is equal to:

$(sin\alpha+cosec \alpha)^2+(cos\alpha+sec\alpha)^2-(tan^2\alpha+cot^2\alpha)$

0%
$0$
0%
$2$
0%
$4$
0%
$7$

Explanation

$(\sin\alpha+cosec \alpha)^2+(\cos\alpha+\sec\alpha)^2-(\tan^2\alpha+\cot^2\alpha)$

$=\sin^2\alpha+cosec^2\alpha+2+\cos^2\alpha+\sec^2\alpha+2-\tan^2\alpha-\cot^2\alpha$

$=4+\sin^2\alpha+\cos^2\alpha+\sec^2\alpha-tan^2\alpha+cosec^2\alpha-\cot^2\alpha=7$

$0 < A < \pi /2$ the value of the expression

$\dfrac {tan A}{1-cot A}+\dfrac {cot A}{1-tan A}-sec A cosec A$ is equal to

0%
-1
0%
0
0%
1
0%
sin A + cos A

Explanation

$\displaystyle \frac {\tan A}{1-\cot A}+\frac {\cot A}{1-\tan A}-\sec A cosec A$

$\displaystyle =\frac {\sin^2A}{\cos A(\sin A-\cos A)}+\frac {\cos^2A}{\sin A(\cos A-\sin A)}-\sec A cosec A$

$\displaystyle =\frac {\sin^3A-\cos^3A}{\sin A\cos A(\sin A-\cos A)}-\sec A cosec A$

$=\displaystyle \frac {\sin^2 A+\cos^2 A+\sin A \cos A}{\sin A \cos A}-\sec A cosec A$

$[a^3 -b^3=(a-b)(a^2+ab+b^2)]$

$=\sec A cosec A+1-\sec A cosec A$

$=1$ .

$\displaystyle \tan A=1$ and

$\displaystyle \tan B=\sqrt{3}$ evaluate:

$\displaystyle \cos A\cos B-\sin A\sin B$

0%
$\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}$
0%
$\displaystyle \frac{\sqrt{2}-\sqrt{6}}{6}$
0%
$\displaystyle \frac{1-\sqrt{3}}{2\sqrt2}$
0%
$\displaystyle \frac{\sqrt{2}+\sqrt{6}}{6}$

Explanation

$\tan A = 1$ and

$\tan B = \sqrt{3}$

$\tan A = \tan 45$ and

$\tan B = \tan 60$

$A = 45^{\circ}$ and

$B = 60^{\circ}$

Now,

$\cos A \cos B - \sin A \sin B$

$\cos 45^0 \cos 60^0 - \sin 45^0 \sin 60^0$

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

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

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

$cos x+sin x=\sqrt 2 cos x$ , then

$tan^2x+2 tan x$ is equal to

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

Explanation

Given

$cos x+sin x=\sqrt 2 cos x$

$(cos x+sin x)^2=2 cos^2 x$

$cos^2x+sin^2x+2sinxcosx=2cos^2x$

Divide

$cos^2x$ both sides

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

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

$\sec C$ is

$\displaystyle \frac{m}{2\sqrt{2}}$ , then

$m$ is:

0%
1
0%
3
0%
7
0%
5

Explanation

Given,

$\triangle ABC$ , A perpendicular from B on AC, let it cut AC at D, such that

$\angle BDA = \angle BDC = 90^{\circ}$

$AD = 3$

$BD = 4$

$BC = 12$

In

$\triangle ABD$ ,

$AB^2 = BD^2 + AD^2$

$AB^2 = 3^2 + 4^2$

$AB = 5$

In

$\triangle BCD$

$BC^2 = CD^2 + BD^2$

$12^2 = CD^2 + 4^2$

$CD^2 = 128$

$CD = 8 \sqrt{2}$

Now,

$\sec C = \frac{H}{P} = \frac{12}{8 \sqrt{2}} = \frac{3}{2\sqrt{2}}$

In the following figure, cosec A is

$\displaystyle \frac{5}{m}$ , m is

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

Explanation

Given,

$\triangle ABC$ , A perpendicular from B on AC, let it cut AC at D, such that

$\angle BDA = \angle BDC = 90^{\circ}$

$AD = 3$

$BD = 4$

$BC = 12$

In

$\triangle ABD$ ,

$AB^2 = BD^2 + AD^2$

$AB^2 = 3^2 + 4^2$

$AB = 5$

In

$\triangle BCD$

$BC^2 = CD^2 + BD^2$

$12^2 = CD^2 + 4^2$

$CD^2 = 128$

$CD = 8 \sqrt{2}$

Now,

$cosec A = \frac{H}{P} = \frac{AB}{BD} = \frac{5}{4}$

$\therefore m=4$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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