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This mock test of NTSE Level Test: Introduction To Trigonometry for Class 10 helps you for every Class 10 entrance exam.
This contains 10 Multiple Choice Questions for Class 10 NTSE Level Test: Introduction To Trigonometry (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The value of (sin 30° + cos 30°) - (sin 60° + cos 60°) is

Solution:

sin 30° = 1/2,

cos 30°=√3/2,

sin 60°=√3/2,

cos 60°=1/2,

By putting the value of sin 30°, cos 30°, sin 60° and cos 60° in equation

We get=

(sin30°+cos30°)-(sin60°+cos60°)=(1/2+√3/2)-(√3/2+1/2)

=0

QUESTION: 2

Ratios of sides of a right triangle with respect to its acute angles are known as

Solution:

The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle.

QUESTION: 3

The value of (sin 45° + cos 45°) is

Solution:

sin 45° + cos 45°

Hence, the answer is = √2

QUESTION: 4

If tan θ = a/b then the value of

Solution:

Let,angle= θ

(asinθ + bcosθ)/(asinθ - bcosθ)

Dividing both numerator and denominator from cosθ

We get,

atanθ +b/atanθ - b

= ( a.a/b + b) /(a.a/b - b) =(a²/b +b)/(a²/b - b)

=(a² + b²/a²- b²)

QUESTION: 5

If 6cotθ + 2cosecθ = cotθ + 5cosecθ, then cosθ is

Solution:

6cot+2cosec=cot+5cosec

6cot-cot=5cosec-2cosec

5cot=3cosec

5cos/sin=3/sin

cos=3/5

QUESTION: 6

Match the Columns:

Solution:

**Correct Answer :- b**

**Explanation :** If θ is one of the acute angles in a triangle, then the sine of theta is the ratio of the opposite side to the hypotenuse, the cosine is the ratio of the adjacent side to the hypotenuse, and the tangent is the ratio of the opposite side to the adjacent side.

QUESTION: 7

9 sec^{2} A - 9tan^{2} A is equal to

Solution:

9 sec^{2} A - 9 tan^{2} A

= 9( sec^{2} A - tan^{2} A)

= 9 × 1

= 9

QUESTION: 8

The value of sin^{2} 30° - cos^{2} 30° is

Solution:

**Solution :- sin^2 30° − cos^2 30°**

**= (1/2) ^{2} −((3)^{1/2}/2)^{2}**

**= 1/4 - 3/4**

**= -1/2**

QUESTION: 9

If tan A = 3/2, then the value of cos A is

Solution:

**Tanθ = Perpendicular / Base**

We are given that TanA = 3/2

On comparing

Perpendicular = 3

Base = 2

To fing hypotenuse

Hypotenuse^{2} = Perpendicular^{2} + Base^{2}

Hypotenuse^{2}^{ }= 3^{2} + 2^{2}

Hypotenuse =

Hypotenuse = 3.6

**Cosθ = Base / Hypotenuse**

CosA = 2 / 3.6

Hence the value of Cos A is 2/3.6=2/√13

QUESTION: 10

If 3 cot θ = 2, then the value of tan θ

Solution:

3 cot θ = 2 ⇒ cot θ = 2/3 ⇒ tan θ = 3/2

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