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Test Level 2: Trigonometry - 1 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 2: Trigonometry - 1

Test Level 2: Trigonometry - 1 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 2: Trigonometry - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 2: Trigonometry - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 2: Trigonometry - 1 below.
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Test Level 2: Trigonometry - 1 - Question 1

Find the value of (cosθ cos(90 - θ) secθ) ÷ (tanθ sin (90 - θ)).

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 1

(cosθ cos(90 - θ) secθ) ÷ (tanθ sin (90 - θ))
= cosθ sinθ secθ ÷ tanθ cosθ
= cosθ sinθ secθ ÷ cosθ sinθ secθ
= 1 

Test Level 2: Trigonometry - 1 - Question 2

If a = x sinθ + y cosθ and b = y sinθ - x cosθ, then y2 - a2 is equal to

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 2

a = x sinθ+ y cosθ
b = y sinθ- x cosθ
y2 - a2 = y2 - (x sinθ+ y cosθ)2
= y2 - x2 sin2θ - y2 cos2θ - 2xy sinθ cosθ
= y2 sin2θ - x2 sin2θ - 2xy sinθ cosθ
= y2 sin2θ + x2 cos2θ - x2 - 2xy sinθ cosθ
= (y sinθ - x cosθ)2 - x2
= b2 - x2

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Test Level 2: Trigonometry - 1 - Question 3

If sin (π cos x) = cos (π sin x), then x equals

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 3

Given: sin (π cos x) = cos (π sin x)

Test Level 2: Trigonometry - 1 - Question 4

In the given figure, AB is the height of a tree. C and D are the two points in a straight line on the ground at b units and a units respectively away from the tree. What is the height of the tree (AB)?

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 4

In ΔABC
AB/BC = tan 60°
AB = √3b units
In Δ ABD
AB/BD = tan 30°
AB = a/√3 units
Now AB × AB = √3b × a/√3 units2 
AB2 = ab units2
AB = √ab units

Test Level 2: Trigonometry - 1 - Question 5

A man on the top of a tower observes a car moving towards the base of the tower at an angle of depression α. Ten minutes later, the angle of depression of the car is found to be β. If the tangent of angle α. is equal to 1/√5, co-tangent of angle β is equal to 1/√5 and the car is moving with a uniform speed, find the total time taken by the car to reach the base of the tower.

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 5


Distance travelled in 10 minutes = CD = x metres

Solving (i) and (ii), we get

CB + x = 5CB
x = 4CB
or, CB = x/4 metres
Total time taken to cover a distance of x/4 metres 

Test Level 2: Trigonometry - 1 - Question 6

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 6



Multiply and divide by 2,

Test Level 2: Trigonometry - 1 - Question 7

If  then the value of sin is

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 7



The componendo - dividendo method state that
Using above theorem in equation (1) we get

Test Level 2: Trigonometry - 1 - Question 8

The angles of elevation of the top of the tower, as observed from each of the points A, B and C on the ground, form a triangle at the same angle α. If R is the circum-radius of the triangle ABC, then the height of the tower is

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 8

The tower makes equal angles at the vertices of the triangle.
Therefore, foot of the tower is at the circumcentre.
From ΔOAP, we have tanα = OP/OA
⇒ OP = OA tan α
⇒ OP = R tan α

Test Level 2: Trigonometry - 1 - Question 9

If sin x + cos y = a and cos x + sin y = b, then what will be the value of tan (x - y)/2 ?

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 9


Test Level 2: Trigonometry - 1 - Question 10

A vertical building and a tower are on the same ground level. From the top of the building, the angle of elevation of the top of the tower is 45° and the angle of depression of the foot of the tower is 60°. Find the height of the tower, if the height of the building is 30 m.

Detailed Solution for Test Level 2: Trigonometry - 1 - Question 10

In the diagram, BC = 30 m = DE, ∠ABD = 45° and ∠EBD = 60°.

Let AD = x m and AE = (x + 30) m.
In right-angled ΔABD,
tan 45° = AD/BD
⇒ 1 = x/BD ⇒ BD = x m … (1)
In right-angled ΔEBD,
tan 60° = DE/BD

From equations (1) and (2), we get

∴ Height of the tower = AE = (x + 30) m
= (10√ + 30) m
= (10 × 1.732 + 30) m = 47.32 m

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