RD Sharma Test: Introduction to Trigonometry - Class 10 MCQ

# RD Sharma Test: Introduction to Trigonometry - Class 10 MCQ

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## 25 Questions MCQ Test Mathematics (Maths) Class 10 - RD Sharma Test: Introduction to Trigonometry

RD Sharma Test: Introduction to Trigonometry for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The RD Sharma Test: Introduction to Trigonometry questions and answers have been prepared according to the Class 10 exam syllabus.The RD Sharma Test: Introduction to Trigonometry MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RD Sharma Test: Introduction to Trigonometry below.
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RD Sharma Test: Introduction to Trigonometry - Question 1

### In ΔABC, ∠B = 90°. If AB = 14 cm and AC = 50 cm then tan A equals :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 1

RD Sharma Test: Introduction to Trigonometry - Question 2

### In sin 3θ = cos (θ – 26°), where 3θ and (θ – 26°) are acute angles, then value of θ is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 2

sin3θ = cos(θ - 26°)

=> cos(90° - 3θ) = cos(θ - 26°)

=> 90° - 3θ = θ - 26°

=> 3θ + θ = 90° + 26°

=> 4θ = 116°

=> θ = 116°/4

=> θ = 29°

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RD Sharma Test: Introduction to Trigonometry - Question 3

### If  sec θ =  then the value of the

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 3

We have,
Dividing by cosθ

Substituting the value of tan θ

RD Sharma Test: Introduction to Trigonometry - Question 4

If angle A is acute and cos A = 8/17 then cot A is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 4

Cos A=8/17=B/H
base=8x, hypotenuse=17x
By pythagoras theorem,
H=P+ B2
289x= P+ 64x2

Cot A=B/P=8x/15x=8/15

RD Sharma Test: Introduction to Trigonometry - Question 5

sec θ is equal to –

RD Sharma Test: Introduction to Trigonometry - Question 6

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 6

It is given that,
∠POQ =110°
Since, the Iangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = 90°
and ∠OQT = 90°
∠POQ + ∠OQT + ∠PTQ + ∠OPT = 360°
⇒ 110° + 90° + ∠PTQ + 90° = 360°
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 70°
Hence, right option is (B).

RD Sharma Test: Introduction to Trigonometry - Question 7

The value of 2 tan2 60° – 4 cos2 45° – 3 sec2 30° is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 7

Step-by-step explanation:

2tan2 60° - 4 cos2 45° -3sec2 30° ----(1)

tan 60° = √3

cos 45° = 1/√2

sec 30° = 2/√3

putting value in equation (1)

2(√3)2 - 4(1/√2)2 - 3(2/√3)2

=2(3) - 4(1/2) - 3(4/3)

=6-2-4

=6-6

=0 ANS

RD Sharma Test: Introduction to Trigonometry - Question 8

The value of 3/4 tan2 30° – 3 sin2 60° + cosec2 45° is

RD Sharma Test: Introduction to Trigonometry - Question 9

7 sin2 θ + 3 cos2 θ = 4 then :

RD Sharma Test: Introduction to Trigonometry - Question 10

The solution of the trigonometric equation

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 10

cos2θ = 3(cot2θ-cos2θ)
4cos2θ = 3cot2θ
4=3(1/sin2θ)
sin2θ = 3/4
sinθ = √3/2
θ = 60

RD Sharma Test: Introduction to Trigonometry - Question 11

In sin 3θ = cos (θ – 26°), where 3θ and (θ – 26°) are acute angles, then value of θ is :

RD Sharma Test: Introduction to Trigonometry - Question 12

The value of sin2 15° + sin2 30° + sin2 45° + sin2 60° + sin2 75° is :

RD Sharma Test: Introduction to Trigonometry - Question 13

The value of  is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 13

RD Sharma Test: Introduction to Trigonometry - Question 14

The values of x and y which make the following solutions true are: cos x° = sin 52° and cos y° = sin (y° + 10)

RD Sharma Test: Introduction to Trigonometry - Question 15

If α + β = 90° and α = 2β then cos2 α + sin2 β equal :

RD Sharma Test: Introduction to Trigonometry - Question 16

A flagstaff 6 metres high throws a shadow 2 √3 metres long on the ground. The angle of elevation is :

RD Sharma Test: Introduction to Trigonometry - Question 17

An observer √3 m tall is 3 m away from the pole 2 √3 m high. The angle of elevation of the top from the pole is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 17

RD Sharma Test: Introduction to Trigonometry - Question 18

An observer 1.5 m tall is 28.5 m away from.a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. The height of the chimney is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 18

RD Sharma Test: Introduction to Trigonometry - Question 19

The angle of elevation of the top of a tower from a distance 100 m from its foot is 60°. The height of the tower is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 19

RD Sharma Test: Introduction to Trigonometry - Question 20

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. The length of the string is:

RD Sharma Test: Introduction to Trigonometry - Question 21

A tree is broken by the wind. Its top struck the ground at an angle 30° at a distance of 30 m from its foot. The whole height of the tree is :

RD Sharma Test: Introduction to Trigonometry - Question 22

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks then the width of the river is :

RD Sharma Test: Introduction to Trigonometry - Question 23

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is:

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 23

Given AB is the tower.

P and Q are the points at distance of 4m and 9m respectively.

From fig, PB = 4m, QB = 9m.

Let angle of elevation from P be α and angle of elevation from Q be β.

Given that α and β are supplementary. Thus, α + β = 90

In triangle ABP,

tan α = AB/BP – (i)

In triangle ABQ,

tan β = AB/BQ

tan (90 – α) = AB/BQ (Since, α + β = 90)

cot α = AB/BQ

1/tan α = AB/BQ

So, tan α = BQ/AB – (ii)

From (i) and (ii)

AB/BP = BQ/AB

AB^2 = BQ x BP

AB^2 = 4 x 9

AB^2 = 36

Therefore, AB = 6.

Hence, height of tower is 6m.

RD Sharma Test: Introduction to Trigonometry - Question 24

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angles of elevation from his eyes to the top of the building increases from 30 to 60° as he walks towards the building. The distance he walked towards the building is :

RD Sharma Test: Introduction to Trigonometry - Question 25

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 60°. If one strip is exactly behind the other on the same side of the light-house then the distance between the two ships is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 25

Height of lighthouse =75m
Angles are 30 and 60
Let x be the distance between the ships and y be the distance between the foot of the lighthouse and closer ship.
So tan 60=

Tan 30 =
x =

## Mathematics (Maths) Class 10

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## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests