MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 9

$\displaystyle25\sin ^{2}\theta +10\cos ^{2}\theta =15$ then find

$\displaystyle\cot ^{2}\theta$

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

Explanation

$25\sin ^{ 2 }{ \theta } +10\cos ^{ 2 }{ \theta } = 15$

$\Rightarrow 15\sin ^{ 2 }{ \theta } + 10\sin ^{ 2 }{ \theta } + 10\cos ^{ 2 }{ \theta } = 15$

$\Rightarrow 15\sin ^{ 2 }{ \theta } + 10(\sin ^{ 2 }{ \theta } + \cos ^{ 2 }{ \theta } ) = 15$

$\Rightarrow 15\sin ^{ 2 }{ \theta } +10 = 15$ (Since,

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

$\Rightarrow 15\sin ^{ 2 }{ \theta } = 5$

$\Rightarrow \sin ^{ 2 }{ \theta } = \dfrac {5}{15} = \dfrac {1}{3}$

And,

$\text{cosec}\, ^{ 2 }{ \theta } = \dfrac {1}{ \sin ^{ 2 }{ \theta }} = 3$

Also,

$\cot ^{ 2 }{ \theta } = \text{cosec}\, ^{ 2 }{ \theta }-1 = 3 - 1 = 2$

In a right angled

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

$\angle A=30^o$ .

Then,

$\cot { 60^o } =$

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

Explanation

Consider an equilateral

$\triangle ABC$ , with each side

$= 2a$ .
So each angle of

$\triangle ABD$ is

$60^o$ .

From

$A$ , draw

$AD \perp BC$ , so

$BD = DC = a$ .

Also,

$\angle BAD=30^o$ and

$\angle ADB=90^o$ .

From right angled

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

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

$\cot { 60^o } =\dfrac { base}{ perpendicular } =\dfrac{BD}{AD}=\dfrac { a }{ \sqrt { 3 } a }=\dfrac{1}{\sqrt3}$ .

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

In a right angle

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

$\angle A=30^o$ .

Then

$\sec { 60 }$ is equal to:

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

Explanation

Consider an equilateral

$\triangle ABC$ , where each side

$= 2a$ .
From

$A$ , draw

$AD \perp BC$ ,

$BD = DC = a$ , and

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

$\triangle ADB$ , we have

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

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

$\sec { 60^o } =\dfrac{hypotenuse}{base}=\dfrac{AB}{BD}=\dfrac{2a}{a}=2$ .

So, option A that is

$2$ is correct.

In right angle

$\triangle ABC, \angle B=90^o, \angle A=45^o$ , then

$\sin { 45^o } =$

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

Explanation

Given

$\angle A=45^o, \angle B=90^o$

$\therefore \angle C=90^o$

$\angle A = \angle C \Rightarrow AB = BC = a$

$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$

$\sin{45^o} = \dfrac{BC}{AC} = \dfrac{a}{\sqrt{2}a} = \dfrac{1}{\sqrt{2}}$

So,

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

In a right angled

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

$\angle A=30^o$ .

Then

$\cos { 30^o }$ 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 of

$\triangle ABC=60^o$ .

From

$A$ draw

$AD \perp BC$ , so

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

From right angled

$\triangle ADB$ , we have

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

$\cos { 30^o } =\dfrac { AD }{ AB } =\dfrac { \sqrt { 3 } a }{ 2a } =\dfrac { \sqrt { 3 } }{ 2 }$ .

So,

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

In a right angled

$\triangle ABD, \angle B=60^o, \angle A=30^o$ .

Then

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

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

Explanation

Consider an equilateral

$\triangle ABC$ , with each side

$= 2a$ .
So, each angle of

$\triangle ABC=60^o$ .

From

$A$ draw

$AD \perp BC$ . So

$BD = DC = a$ ,

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

From right angled

$\triangle ADB$ , we have

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

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

$\sin { 30^o } =\dfrac { BD }{ AB } =\dfrac { a }{ 2a } =\dfrac { 1 }{ 2 }$ .

So,

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

In right angled

$\triangle ABD, \angle B=60^o, \angle A=30^o$ .

Then

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

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

Explanation

Consider an equilateral

$\triangle ABC$ , with each side

$= 2a$ .

So, each angle is

$60^o$ .

From

$A$ , draw

$AD \perp BC$ .

So,

$BD=DC=a, \angle BAD=30^o$ and

$\angle ADB=90^o$ .

So, for right angled

$\triangle ADB$ , we have

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

$\sec { 30^o } =\dfrac { 1 }{ \cos { 30^o } } =\dfrac { 1 }{ \dfrac { \sqrt { 3 } }{ 2 } } =\dfrac{AB}{AD}=\dfrac { 2 }{ \sqrt { 3 } }$ .
So

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

In a right angled

$\triangle ABD$ , in which

$\angle B=60^o$ and

$\angle A=30^o$ .

Then

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

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

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

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

Now

$\csc { 30 } =\dfrac {AB}{BD} =\dfrac { 2a }{ a } =2$ .

In right angle

$\triangle ABD, \angle B={ 60 }^{ o }, \angle A={ 30 }^{ o }$ , then

$\cot { { 60 }^{ o } } =$

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

Explanation

$\triangle ABD$

$\cot60^o=\cot B=\dfrac{base}{perpendicular}=\dfrac{a}{\sqrt3a}=\dfrac{1}{\sqrt3}$

Therefore, Answer is

$\dfrac{1}{\sqrt3}$

In right angle

$\triangle ABC, \angle B=90^o, \angle A=45^o$ , then

$\cos { 45^o } =$

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

Explanation

Given

$\angle A=45^o, \angle B=90^o$

$\therefore \angle C=45^o$

$\angle A = \angle C \Rightarrow AB = BC$

Let

$AB=BC=a$

$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$

$\cos{45} = \dfrac{AB}{AC} = \dfrac{a}{\sqrt{2}a} = \dfrac{1}{\sqrt{2}}$

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

Value of

$\left(\sin{60^o}\cos{30^o}-\cos{60^o}\sin{30^o}\right)$ is

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

Explanation

Given that:

$\left( \sin { 60^o } \cos { 30^o } -\cos { 60^o } \sin { 30^o } \right)$

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

$=\left( \dfrac { 3 }{ 4 } -\dfrac { 1 }{ 4 } \right)$

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

In right angle

$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$ , then

$cosec { 45^o } =$

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

Explanation

$\angle A = \angle C \Rightarrow AB = BC = a$

$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$

$\csc{45^o} = \dfrac{hypotenuse}{base} =\dfrac{AC}{AB}= \dfrac{\sqrt2 a}{{a}} = \sqrt{2}$

So,

$\sqrt{2}$ is correct.

In right angle

$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$ , then

$\cot { 45^o } =$

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

Explanation

$\angle A = \angle C \Rightarrow AB = BC = a$

$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$

$\cos{45^o} = \dfrac{base}{perpendicular} =\dfrac{AB}{BC}= \dfrac{1}{1} = 1$

$1$ is correct.

In right angle

$\triangle ABC, \angle B=90^o, \angle A=45^o$ , then

$\tan { 45^o } =$

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

Explanation

$\angle A = \angle C \Rightarrow AB = BC = a$

$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$

$\tan{45} = \dfrac{BC}{AB} = \dfrac{a}{a} =1$

$1$ is correct.

In right angle

$\triangle ABC, \angle B=90^o, \angle A=45^o=\angle C$ , then

$\sec { 45^o} =$

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

Explanation

$\angle A = \angle C \Rightarrow AB = BC = a$

$AC = \sqrt{{AB}^{2}+{BC}^{2}} = \sqrt{{a}^{2}+{a}^{2}} = \sqrt{2{a}^{2}} = \sqrt{2} a$

$\sec{45^o} = \dfrac{hypotenuse}{base} =\dfrac{AC}{AB}= \dfrac{\sqrt2a}{a} = \sqrt{2}$
So,

$\sqrt{2}$ is correct.

${\tan}^{2}{45^o}- {\cos}^{2}{30^o} = x \sin{45^o} \cos{45^o}$ , then

$x =$

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

Explanation

Given that:

$\tan ^{ 2 }{ 45^o } -\cos ^{ 2 }{ 30^o } =x\sin { 45^o } \cos { 45^o }$

$x=\dfrac { \tan ^{ 2 }{ 45^o } -\cos ^{ 2 }{ 30^o } }{ \sin { 45^o } \cos { 45^o } }$

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

Value of

$\left(\cos{60^o}\cos{30^o}-\sin{60^o}\sin{30^o}\right)$ is

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

Explanation

$\left( \cos { 60^0 } \cos { 30^0-\sin { 60^0 } \sin { 30^0 } } \right)$

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

$=\dfrac { \sqrt { 3 } }{ 4 } -\dfrac { \sqrt { 3 } }{ 4 }$

$=0$

$\triangle ABC, \angle B=90^o, AB=3\ cm, AC=6\ cm, \angle A=$

0%
$30^o$
0%
$45^o$
0%
$60^o$
0%
None of these

Explanation

$\sin{C} = \dfrac{AB}{AC} = \dfrac{3}{6} = \dfrac{1}{2}$

But

$\sin{30^o} = \dfrac{1}{2}$ , so,

$\angle C = 30^o$

$\angle A = \left(90-\angle C\right)^o = \left(90-30\right)^o = 60^o$ .

$60^o$ is correct.

$\triangle ABC, \angle B=90^o, \angle A=30^o, AB =9 cm\$ , then

$BC =$

0%
$3 cm$
0%
$2\sqrt{3} cm$
0%
$3\sqrt{3} cm$
0%
$6 cm$

Explanation

$\tan{30^0} = \dfrac{BC}{AB}$ and

$\tan{30^0} = \dfrac{1}{\sqrt{3}}$

$\therefore \dfrac{BC}{9} = \dfrac{1}{\sqrt{3}}$ , so

$BC = \dfrac{9}{\sqrt{3}} = 3\sqrt{3}$ .

So,

$3\sqrt{3}$ is correct.

$\sec (90 - \theta) =$

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

Explanation

In the figure,

$\sec (90 - \theta) = \dfrac {r}{y} = \csc \theta$

So,

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

So,

$\csc \theta$ is correct.

$(\cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ}) (\sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ}) =$

0%
$\dfrac {5}{4}$
0%
$\dfrac {9}{4}$
0%
$\dfrac {7}{4}$
0%
None of these

Explanation

$\left( \cos { { 0^o } + } \sin { { 45^o } + } \sin { { 30^o } } \right) \left( \sin { { 90^o- } } \cos { { 45^o+ } } \cos {{ 60^o } } \right)$

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

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

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

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

$=\dfrac { 9 }{ 4 } -\dfrac { 1 }{ 2 } =\dfrac { 7 }{ 4 }$

$\cot (90 - \theta) =$

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

Explanation

$\cot (90 - \theta) = \dfrac {y}{x} = \tan \theta$

So,

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

So,

$\tan \theta$ is correct.

$A$ and

$B$ are acute angle such that

$\sin A = \cos B$ , then

$(A + B) = .....$

0%
$45^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$
0%
$180^{\circ}$

Explanation

$\sin A=\cos B$

$\Rightarrow \sin A=\sin(90^o-B)$

$\Rightarrow A=90^o-B$

$\Rightarrow A+B=90^o$

Therefore, Answer is

$90^o$

$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 } \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 } \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 } =$

0%
$\dfrac { 8 }{ 3 }$
0%
$\dfrac { 11 }{ 6 }$
0%
$\dfrac { 13 }{ 6 }$
0%
$\dfrac { 13 }{ 8 }$

Explanation

$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 } \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 } \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 }$

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

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

$=4\times \dfrac { 1 }{ 8 } -\dfrac { 2 }{ 3 } \times \dfrac { 1 }{ 4 } +\dfrac { 3 }{ 2 }$

$=\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 6 } +\dfrac { 3 }{ 2 }$

$=\dfrac { 11 }{ 6 }$

$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚ =$

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

Explanation

$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚$

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

$= 1 + 2 - \dfrac {3}{2} - \dfrac {3}{2}$

$= 0$

$\sin 30^{\circ} \cos 60^{\circ} + \sin 60^{\circ} \cos 30^{\circ}$ is equal to

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

Explanation

Given:

$\sin { { 30^o } } \cos { { 60^o } } +\sin { { 60 ^o } } \cos { { 30^o } }$

(Putting the values as per trigonometry table)

$=\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 }$

$=\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } =\dfrac { 4 }{ 4 } =1$

$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 } \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 } \right] -5\cos ^{ 2 }{ 60^0 } =$

0%
$\dfrac { 35 }{ 9 }$
0%
$\dfrac { 45 }{ 8 }$
0%
$\dfrac { 55 }{ 18 }$
0%
$\dfrac { 49 }{ 36 }$

Explanation

$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 } \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 } \right] -5\cos ^{ 2 }{ 60^0 } =$

$=\dfrac { 1 }{ 4 } \left[ { \left( \sqrt { 3 } \right) }^{ 4 }-{ \left( \dfrac { 2 }{ \sqrt { 3 } } \right) }^{ 4 } \right] +\dfrac { 3 }{ 2 } \left[ { \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 } \right] -5\times { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$ , substitute the values from the table

$=\dfrac { 1 }{ 4 } \left( 9-\dfrac { 16 }{ 9 } \right) +\dfrac { 3 }{ 2 } \left( 2-\dfrac { 1 }{ 3 } \right) -\dfrac { 5 }{ 4 }$

$=\dfrac { 65 }{ 36 } +\dfrac { 5 }{ 2 } -\dfrac { 5 }{ 4 } =\dfrac { 110 }{ 36 } =\dfrac { 55 }{ 18 }$

$\tan (90 - \theta)$ is equivalent to

0%
$\cos \theta$
0%
$\sin \theta$
0%
$(\sin \theta + \cos \theta)$
0%
$\cot \theta$

Explanation

In the figure,

$\tan (90^{\circ} - \theta) = \dfrac {x}{y}$ and

$\cot \theta = \dfrac {x}{y}$

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

$\cot \theta$ is correct.

$\triangle ABC$ , if

$AD\perp BC$ and

$BD = 10\ cm; \angle B = 60^{\circ}$ and

$\angle C = 30^{\circ}$ , then

$CD = .........$

0%
$10\ cm$
0%
$20\ cm$
0%
$40\ cm$
0%
$30\ cm$

Explanation

In right angled

$\triangle ABD$ ,

$\tan B = \dfrac {AD}{BD} = \dfrac {AD}{10}$

$\Rightarrow \tan 60^{\circ} = \dfrac {AD}{10}\Rightarrow \sqrt {3} = \dfrac {AD}{10}$

$\Rightarrow AD = 10\sqrt {3}$

Also, in

$\triangle ACD$ ,

$\tan 30^{\circ} = \dfrac {AD}{DC} = \dfrac {1}{\sqrt {3}} \Rightarrow \dfrac {10\sqrt {3}}{DC} = \dfrac {1}{\sqrt {3}}$

$\Rightarrow DC = 30\ cm$ .

$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0 =$

0%
$\dfrac {1}{2}$
0%
$\dfrac {3}{2}$
0%
$\dfrac {5}{2}$
0%
$\dfrac {7}{2}$

Explanation

Taking

$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0$

Substituting the values of sin

$30 ^0$ , sin

$60 ^0$ , cos

$45 ^0$

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

$=\dfrac { 1 }{ 2 } +\dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 }$

$=\dfrac { 6 }{ 4}$

$=\dfrac { 3 }{ 2 }$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 9 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 9 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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