Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

If $$\sec{2A}=\csc{(A-42^\circ)}$$ where $$2A$$ is acute angle then value of $$A$$ is

A
$$44^\circ$$

B
$$22^\circ$$

C
$$21^\circ$$

D
$$66^\circ$$

Solution

$$\sec{2A}=\csc{(A-42^\circ)}$$
$$\csc{(90-2A)}=\csc{(A-42^\circ)}$$
$$90-2A=A-42$$
$$3A=132$$
$$A=44^\circ$$
Question 2
1 / -0

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

A
$$\displaystyle\frac{1}{\sqrt{2}}$$

B
0

C
1

D
$$\displaystyle\frac{1}{2}$$

Solution

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

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

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

$$=1-\dfrac 1 2$$

$$=\dfrac 1 2$$
Question 3
1 / -0

If $$\displaystyle 2\sin 2\theta =\sqrt{3}$$ then the value of $$\displaystyle \theta$$ is

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

B
$$\displaystyle 45^{\circ}$$

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

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

Solution

$$2sin 2\theta=\sqrt{3}$$
$$\Rightarrow sin 2\theta=\dfrac{\sqrt{3}}{2}$$
$$\because sin60^\circ{}=\dfrac{\sqrt{3}}{2}$$
$$\therefore sin 2\theta=sin 60^\circ{}$$
$$\Rightarrow 2\theta=60^\circ{}$$
$$\Rightarrow \theta=\dfrac{60}{2}=30^\circ{}$$
Question 4
1 / -0

For an acute angle $$\theta$$ in a right angled triangle, $$\dfrac{1}{\text{cosec}\,{\theta}} =$$

A
$$\sin{\theta}$$

B
$$\cos{\theta}$$

C
$$\tan{\theta}$$

D
None of these

Solution

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

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

$$\text{cosec}\ \theta = \dfrac{hypotenuse}{opposite}$$
$$\dfrac{1}{\text{cosec}\ \theta}=\dfrac{1}{\dfrac{hypotenuse}{opposite}}=\dfrac{opposite}{hypotenuse}=\sin\ \theta$$
$$\therefore\ \dfrac{1}{\text{cosec}\ \theta}=\sin\ \theta$$
Question 5
1 / -0

The value of $$\cos0^0$$ is ______.

A
$$0$$

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

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

D
$$1$$

Solution

Using the trigonometric table
The value of $$\cos { 0^\circ } $$ is 1
Question 6
1 / -0

$${\sec}^{2}{60^o} - 1= $$

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

To find:
$$\sec ^{ 2 }{ { 60^o-1 } } $$
$$={ \left( 2 \right) }^{ 2 }-1$$
$$=4-1=3$$
Question 7
1 / -0

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

A
$$0$$

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

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

D
$$1$$

Solution

Given that: $$\left( \cos { 60^o } \cos { 30^o-\sin { 60^o } \sin { 30^o } } \right) $$
$$=\dfrac { 1 }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } -\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { 1 }{ 2 } $$
$$=\dfrac { \sqrt { 3 } }{ 4 } -\dfrac { \sqrt { 3 } }{ 4 } $$
$$=0$$
Question 8
1 / -0

For an acute angle $$\theta$$ in a right angled triangle, $$\dfrac{\sin{\theta}}{\cos{\theta}} =$$

A
$$\cot{\theta}$$

B
$$\cos{\theta}$$

C
$$\sin{\theta}$$

D
$$\tan{\theta}$$

Solution

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

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

$$\dfrac{\sin \theta}{\cos \theta}=\dfrac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}}$$

$$=\dfrac{opposite}{adjacent}\times \dfrac{hypotenuse}{hypotenuse}$$

$$=\dfrac{opposite}{adjacent}$$
$$=\tan \theta$$
Hence, $$\dfrac{\sin \theta}{\cos \theta}=\tan \theta$$.
Question 9
1 / -0

For an acute angle $$\theta$$ in a right angled triangle, $$\dfrac{\cos{\theta}}{\sin{\theta}} =$$

A
$$\cot{\theta}$$

B
$$\sin{\theta}$$

C
$$\cos{\theta}$$

D
$$\tan{\theta}$$

Solution

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

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

$$\dfrac{\cos \theta}{\sin \theta}=\dfrac{\frac{\mathrm adjacent}{\mathrm hypotenuse}}{\frac{\mathrm opposite}{\mathrm hypotenuse}}$$

$$=\dfrac{adjacent}{opposite}\times \dfrac{hypotenuse}{hypotenuse}$$

$$=\dfrac{adjacent}{opposite}$$
$$=\cot \theta$$
Hence, $$\dfrac{\cos \theta}{\sin \theta}=\cot \theta$$.
Question 10
1 / -0

For an acute angle $$\theta$$ in a right angled triangle, $$\dfrac{1}{\tan{\theta}} =$$

A
$$\text{cosec}\,{\theta}$$

B
$$\sec{\theta}$$

C
$$\cot{\theta}$$

D
None of these

Solution

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

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

$$\dfrac{\sin \theta}{\cos \theta}=\dfrac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}}$$
$$=\dfrac{opposite}{adjacent}$$
$$=\tan \theta$$
$$\therefore \dfrac{\sin \theta}{\cos \theta}=\tan \theta$$.

$$\dfrac{1}{\tan \theta}=\dfrac{\cos \theta}{\sin \theta}=\dfrac{\frac{adjacent}{hypotenuse}}{\frac{opposite}{hypotenuse}}$$
$$=\dfrac{adjacent}{opposite}$$
$$=\cot \theta$$
$$\therefore \dfrac{1}{\tan \theta}=\cot \theta$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 21

Result

Introduction to Trigonometry Test - 21

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SHARING IS CARING

Question 1

$$44^\circ$$

$$22^\circ$$

$$21^\circ$$

$$66^\circ$$

Solution

Question 2

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

0

1

$$\displaystyle\frac{1}{2}$$

Solution

Question 3

$$\displaystyle 30^{\circ}$$

$$\displaystyle 45^{\circ}$$

$$\displaystyle 60^{\circ}$$

$$\displaystyle 90^{\circ}$$

Solution

Question 4

$$\sin{\theta}$$

$$\cos{\theta}$$

$$\tan{\theta}$$

None of these

Solution

Question 5

$$0$$

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

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

$$1$$

Solution

Question 6

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 7

$$0$$

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

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

$$1$$

Solution

Question 8

$$\cot{\theta}$$

$$\cos{\theta}$$

$$\sin{\theta}$$

$$\tan{\theta}$$

Solution

Question 9

$$\cot{\theta}$$

$$\sin{\theta}$$

$$\cos{\theta}$$

$$\tan{\theta}$$

Solution

Question 10

$$\text{cosec}\,{\theta}$$

$$\sec{\theta}$$

$$\cot{\theta}$$

None of these

Solution

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