Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 25

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Question 1
1 / -0

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

A
$$\sin 0^\circ$$

B
$$\sin 90^\circ$$

C
$$\cot 45^\circ$$

D
$$\cos 0^\circ$$

Solution

Given: $$\tan \theta =1$$

We observe that

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

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

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

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

Hence $$\sin \; 0^{\circ} \ne \tan \theta$$
Question 2
1 / -0

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

A
$$10$$

B
$$3$$

C
$$7$$

D
$$9$$

Solution

$$\tan 5\theta . \tan4\theta =1$$
$$\therefore \tan 5\theta=\dfrac{1}{\tan4\theta}$$
$$\therefore \tan5 \theta = \cot4 \theta$$
$$\therefore\tan 5\theta =\tan (90^o-4\theta)$$
$$\therefore 5\theta =90^o-4\theta$$
$$\therefore 9\theta=90^o$$
$$\theta =10^o$$
Question 3
1 / -0

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

A
$$1$$

B
$$\cfrac{1}{2}$$

C
$$0$$

D
$$-1$$

Solution

$$\cos {12}^{\circ}= \sin{( 90 - 12)}^{\circ} = \sin{78}^{\circ}$$
So, $$\cos(12^\circ) - \sin (78^\circ) = \sin(78^\circ) - \sin(78^\circ) = 0$$
Question 4
1 / -0

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

A
$$0$$

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

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

D
$$1$$

Solution

$$\sin { 60° } -\cos { 30° } $$
$$=\dfrac { \sqrt { 3 } }{ 2 } -\dfrac { \sqrt { 3 } }{ 2 }$$
$$=0$$
Hence, the answer is $$0.$$
Question 5
1 / -0

If $$2\sin\theta-3\cos\theta=0$$ then $$\tan\theta$$=........ .

A
$$\dfrac{2}{3}$$

B
$$\dfrac{3}{2}$$

C
$$1$$

D
$$\dfrac{1}{2}$$

Solution

Acccording to question ,

$$2sin\theta-3cos\theta=0$$

=> $$2sin\theta=3cos\theta$$

=> $$\dfrac{sin\theta}{cos\theta}=\dfrac{3}{2}$$

=> $$tan\theta=\dfrac{3}{2}$$
Question 6
1 / -0

What is the value of $$\sin{45^{o}}+\cos{45^{o}}$$

A
$$\sqrt {7}$$

B
$$\sqrt {5}$$

C
$$\sqrt {2}$$

D
$$\sqrt {3}$$

Solution

According to question,

$$sin45^o+cos45^o$$

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

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

$$\implies \sqrt2$$
Question 7
1 / -0

From the given figure, AB =

A
16

B
8

C
4

D
12

Solution

In $$ \triangle ADC ,$$

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

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

$$\Rightarrow x = 1 $$

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

In $$ \triangle ABD $$

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

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

$$\Rightarrow AB = 8 $$
Question 8
1 / -0

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

A
$$\cfrac { 7 }{ 3 } $$

B
$$\cfrac { 3 }{ 7 } $$

C
$$\cfrac { -7 }{ 3 } $$

D
$$\cfrac { -3 }{ 7 } $$

Solution

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

We know,

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

Hence, $$\cos \theta = \dfrac{3}{7}$$
Question 9
1 / -0

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

A
$$cos{ 9 }^{ 0 }+cot{ 81 }^{ 0 }$$

B
$$cos{ 9 }^{ 0 }+tan{ 81 }^{ 0 }$$

C
$$cos{ 9 }^{ 0 }+cot{ 9 }^{ 0 }$$

D
$$sin81^{ 0 }+cot{ 81 }^{ 0 }$$

Solution

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

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

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

Hence, option C is correct.
Question 10
1 / -0

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

A
$$1$$

B
$$\sqrt3$$

C
$$0$$

D
can't be determined

Solution

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

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

$$=1$$

So the relation is $$\text{True}$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 25

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Introduction to Trigonometry Test - 25

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Question 1

$$\sin 0^\circ$$

$$\sin 90^\circ$$

$$\cot 45^\circ$$

$$\cos 0^\circ$$

Solution

Question 2

$$10$$

$$3$$

$$7$$

$$9$$

Solution

Question 3

$$1$$

$$\cfrac{1}{2}$$

$$0$$

$$-1$$

Solution

Question 4

$$0$$

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

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

$$1$$

Solution

Question 5

$$\dfrac{2}{3}$$

$$\dfrac{3}{2}$$

$$1$$

$$\dfrac{1}{2}$$

Solution

Question 6

$$\sqrt {7}$$

$$\sqrt {5}$$

$$\sqrt {2}$$

$$\sqrt {3}$$

Solution

Question 7

16

8

4

12

Solution

Question 8

$$\cfrac { 7 }{ 3 } $$

$$\cfrac { 3 }{ 7 } $$

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

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

Solution

Question 9

$$cos{ 9 }^{ 0 }+cot{ 81 }^{ 0 }$$

$$cos{ 9 }^{ 0 }+tan{ 81 }^{ 0 }$$

$$cos{ 9 }^{ 0 }+cot{ 9 }^{ 0 }$$

$$sin81^{ 0 }+cot{ 81 }^{ 0 }$$

Solution

Question 10

$$1$$

$$\sqrt3$$

$$0$$

can't be determined

Solution

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