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RD Sharma Test: Introduction to Trigonometry - Grade 10 MCQ


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25 Questions MCQ Test - RD Sharma Test: Introduction to Trigonometry

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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

3. In triangle ABC.angle B =90°. If AB =14 cm and AC =50 cm then tanA=O  24/2524/7O 7/2425/24​ - Brainly.in

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

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

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 4

sec θ is equal to –

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

As we know,

RD Sharma Test: Introduction to Trigonometry - Question 5

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 5

It is given that,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
∠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°
Now, in quadrilateral POQT.
∠POQ + ∠OQT + ∠PTQ + ∠OPT = 360°
(angle sum property of quadrilateral)
⇒ 110° + 90° + ∠PTQ + 90° = 360°
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 70°
Hence, right option is (B).

RD Sharma Test: Introduction to Trigonometry - Question 6

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

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

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 7

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

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

RD Sharma Test: Introduction to Trigonometry - Question 8

7 sin2 θ + 3 cos2 θ = 4 then :

RD Sharma Test: Introduction to Trigonometry - Question 9

The solution of the trigonometric equation 

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 9


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 10

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 10

RD Sharma Test: Introduction to Trigonometry - Question 11

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

RD Sharma Test: Introduction to Trigonometry - Question 12

The value of  is :

Detailed Solution for RD Sharma Test: Introduction to Trigonometry - Question 12

RD Sharma Test: Introduction to Trigonometry - Question 13

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 14

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

RD Sharma Test: Introduction to Trigonometry - Question 15

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 16

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 16

RD Sharma Test: Introduction to Trigonometry - Question 17

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 17

RD Sharma Test: Introduction to Trigonometry - Question 18

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 18

RD Sharma Test: Introduction to Trigonometry - Question 19

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 20

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 21

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 22

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 22

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 23

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 24

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 24

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 = 

RD Sharma Test: Introduction to Trigonometry - Question 25

 In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is:

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

Answer: (b) 24/7

Sol: AB = 24 cm and BC = 7 cm

tan C = Opposite side/Adjacent side

tan C = 24/7

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