Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

In $$\triangle ABC$$, $$\angle B= {90}^{\circ}$$, $$\sin{C} = \dfrac{3}{5}$$, then $$\cos{A} =$$ ______

A
$$\dfrac{3}{5}$$

B
$$\dfrac{4}{5}$$

C
$$\dfrac{5}{4}$$

D
$$\dfrac{5}{3}$$

Solution

In a right angled triangle, $$A+B+C=180^\circ$$
$$B=90^\circ$$
So, $$A+C=90^\circ$$
$$A=90^\circ-C$$
$$\cos A=\cos(90^\circ-A)$$
$$\Rightarrow \cos A=\sin C=\dfrac35$$
$$\Rightarrow \cos A=\cfrac35$$
Question 2
1 / -0

In the following figure $$AB \perp BC$$ and $$\angle ACB = 30^\circ$$, given $$BC = \sqrt{300}\ m$$. The length of $$AB$$ is

A
$$10\ m$$

B
$$100\ m$$

C
$$10 \sqrt 3\ m$$

D
$$100 \sqrt 3\ m$$

Solution

Given that:
From the fig; $$AB\bot BC,\angle ACB=30^\circ, BC=\sqrt300 m$$
To find:
$$AB=?$$
Solution:
In $$\triangle ABC,$$
$$\tan\angle ACB=\cfrac{AB}{BC}$$
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.
Question 3
1 / -0

If $$\sin \theta = \dfrac {3}{5}$$, then the value of $$\text{cosec}\, \theta$$ is

A
$$\dfrac {4}{5}$$

B
$$\dfrac {5}{3}$$

C
$$\dfrac {4}{3}$$

D
$$\dfrac {5}{4}$$

Solution

Given $$\sin \ \theta=\dfrac{3}{5}$$
We know that $$\text{cosec}\, \theta=\dfrac{1}{\sin \ \theta}=\dfrac{1}{\dfrac{3}{5}}$$
$$\therefore$$ $$\text{cosec}\, \theta=\dfrac{5}{3}$$
Question 4
1 / -0

The value of $$2\sin 30^{\circ} \cos 30^{\circ}$$ is equal to

A
$$\tan 30^{\circ}$$

B
$$\cos 60^{\circ}$$

C
$$\sin 60^{\circ}$$

D
$$\cot 60^{\circ}$$

Solution

$$2\sin { 30° } \cos { 30° }$$

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

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

$$=\sin { 60° } $$

Hence, the answer is $$\sin { 60° }. $$
Question 5
1 / -0

The value of $$\dfrac {\sin 27^{\circ}}{\cos 63^{\circ}}$$ is

A
$$0$$

B
$$1$$

C
$$\tan 27^{\circ}$$

D
$$\cot 63^{\circ}$$

Solution

$$\dfrac { \sin { 27° } }{ \cos { 63° } }$$
$$=\dfrac { \sin { \left( 90°-63° \right) } }{ \cos { 63° } } $$
$$=\dfrac { \cos { 63° } }{ \cos { 63° } }$$
$$=1$$
Hence, the answer is $$1.$$
Question 6
1 / -0

The value of $$cosec^{2} 60^{\circ} - 1$$ is equal to

A
$$\cos^{2} 60^{\circ}$$

B
$$\cot^{2} 60^{\circ}$$

C
$$\sec^{2} 60^{\circ}$$

D
$$\tan^{2} 60^{\circ}$$
Question 7
1 / -0

If $$x = \dfrac {2\tan 30^{\circ}}{1 - \tan^{2} 30^{\circ}}$$, then the value of $$x$$ is:

A
$$\tan 45^{\circ}$$

B
$$\tan 30^{\circ}$$

C
$$\tan 60^{\circ}$$

D
$$\tan 90^{\circ}$$

Solution

$$x=\dfrac { 2\tan { 30° } }{ 1-\tan ^{ 2 }{ 30° } } =\dfrac { 2\times \dfrac { 1 }{ \sqrt { 3 } } }{ 1-{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 } } =\dfrac { \dfrac { 2 }{ \sqrt { 3 } } }{ 1-\dfrac { 1 }{ 3 } } $$
$$\Rightarrow x=\dfrac { 2 }{ \sqrt { 3 } } \times \dfrac { 3 }{ 2 } =\sqrt { 3 } =\tan { 60° } $$
$$\Rightarrow x=\tan { 60° } .$$
Hence, the answer is $$\tan { 60° } .$$
Question 8
1 / -0

If $$\cos x = \sin 43^{\circ}$$, then the value of $$x$$ is

A
$$57^{\circ}$$

B
$$43^{\circ}$$

C
$$47^{\circ}$$

D
$$90^{\circ}$$

Solution

$$\cos { x } =\sin { 43° } $$
$$\Rightarrow \cos { x } =\sin { \left( 90°-47° \right) } $$
$$\Rightarrow \cos { x } =\cos { 47° } $$
$$\Rightarrow x=47°$$
Hence, the answer is $$47°.$$
Question 9
1 / -0

The value of $$\sin^{2} 60^{\circ} + \cos^{2} 60^{\circ}$$ is equal to

A
1

B
2

C
3

D
$$0$$

Solution

$$\sin ^{ 2 }{ 60° } +\cos ^{ 2 }{ 60° } ={ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$$
$$\Rightarrow \dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 } =\dfrac { 4 }{ 4 } =1$$
Question 10
1 / -0

If $$sin^2 (3x + 45) + cos^2 (2x + 60) = 1$$ then $$x =$$ ...............

A
$$60$$

B
$$30$$

C
$$15$$

D
$$0$$

Solution

$$sin^2 (3x + 45)+ cos^2 (2x + 60) = 1$$
$$\therefore sin^2 (3x + 45) = 1 - cos^2 (2x + 60)$$
$$\therefore sin^2 (3x + 45) = sin^2 (2x + 60)$$ ............. [since $$1 - cos^2 \theta = sin^2 \theta$$]
$$\therefore 3x + 45 = 2x + 60$$
$$\therefore x = 15$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 62

Result

Introduction to Trigonometry Test - 62

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SHARING IS CARING

Question 1

$$\dfrac{3}{5}$$

$$\dfrac{4}{5}$$

$$\dfrac{5}{4}$$

$$\dfrac{5}{3}$$

Solution

Question 2

$$10\ m$$

$$100\ m$$

$$10 \sqrt 3\ m$$

$$100 \sqrt 3\ m$$

Solution

Question 3

$$\dfrac {4}{5}$$

$$\dfrac {5}{3}$$

$$\dfrac {4}{3}$$

$$\dfrac {5}{4}$$

Solution

Question 4

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

$$\cos 60^{\circ}$$

$$\sin 60^{\circ}$$

$$\cot 60^{\circ}$$

Solution

Question 5

$$0$$

$$1$$

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

$$\cot 63^{\circ}$$

Solution

Question 6

$$\cos^{2} 60^{\circ}$$

$$\cot^{2} 60^{\circ}$$

$$\sec^{2} 60^{\circ}$$

$$\tan^{2} 60^{\circ}$$

Question 7

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

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

$$\tan 60^{\circ}$$

$$\tan 90^{\circ}$$

Solution

Question 8

$$57^{\circ}$$

$$43^{\circ}$$

$$47^{\circ}$$

$$90^{\circ}$$

Solution

Question 9

1

2

3

$$0$$

Solution

Question 10

$$60$$

$$30$$

$$15$$

$$0$$

Solution

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