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CAT Exam  >  CAT Test  >  Level-wise Tests  >  Test Level 1: Trigonometry - 1 - CAT MCQ

Test Level 1: Trigonometry - 1 CAT MCQs & solutions - Free


MCQ Practice Test & Solutions: Test Level 1: Trigonometry - 1 (10 Questions)

You can prepare effectively for CAT Level-wise Tests for CAT with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test Level 1: Trigonometry - 1". These 10 questions have been designed by the experts with the latest curriculum of CAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 10

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Test Level 1: Trigonometry - 1 - Question 1
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If cot (x + 27°) = tan y, then which of the following equations is true?  

Detailed Solution: Question 1

cot (90 - θ) = tan θ
So, tan y = cot (90° - y)
Hence, cot (x + 27°) = cot (90° - y)
x + 27° = 90° - y
x + y = 90° - 27° = 63°

Test Level 1: Trigonometry - 1 - Question 2
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Solve for x: 2 cos (3x) - 1 = 0; where, x ∈ (0, 2π)

Detailed Solution: Question 2

2 cos 3x - 1 = 0
⇒ cos 3x = 1/2 = cos 60°
The general solution of above equation is 

The solution in the interval (0, 2π) are

Hence, correct answer that match the given options are 20°, 100°

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Test Level 1: Trigonometry - 1 - Question 3
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The value of cos 24° + cos 55° + cos 125° + cos 204° + cos 300° is

Detailed Solution: Question 3

The value of cos 24° + cos 55° + cos 125° + cos 204° + cos 300° ?
We know that, cos(180° ± θ) = - cosθ
⇒ cos 24° + cos 55° + cos(180° - 55°) + cos(180° + 24°) + cos(360° - 60°)
= cos 24° + cos 55° - cos 55° - cos 24° + cos 60°
= cos 60° = 1/2

Test Level 1: Trigonometry - 1 - Question 4
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Solve the equation 2 sin2 x + 3 cos x = 0.

Detailed Solution: Question 4

2 sin2 x + 3 cos x = 0
⇒ 2(1 - cos2 x) + 3 cos x = 0
⇒ 2 - 2 cos2 x + 3 cos x = 0
⇒ 2 cos2 x - 3 cos x - 2 = 0
⇒ 2 cos2 x - 4 cos x + cos x - 2 = 0
⇒ 2 cos x(cos x - 2) + 1(cos x - 2) = 0
⇒ (cos x - 2)(2 cos x + 1) = 0
Either cos x - 2 = 0 or 2 cos x + 1 = 0
But cos x - 2 = 0
i.e. cos x = 2, which is not possible.
Now, from 2 cos x + 1 = 0, we get:
cos x = -1/2

Therefore, general solution of the equation is:

Test Level 1: Trigonometry - 1 - Question 5
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Find the value of cos2 15° - cos2 30° + cos2 45° - cos2 60° + cos2 75°.

Detailed Solution: Question 5

cos2 15° - cos2 30° + cos2 45° - cos2 60° + cos2 75° = (cos2 15° + cos2 75°) - (cos2 30° + cos2 60°) + cos2 45°
= [cos2 15° + cos2 (90° - 15°)] - [cos2 30° + cos2 (90° - 30°)] + cos2 45°
= [cos2 15° + sin2 15°] - [cos2 30° + sin2 30°] + cos2 45° [∵ cos (90° - θ) = sin θ]

Test Level 1: Trigonometry - 1 - Question 6
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Two pillars of equal height stand on either side of a road, which is 150 m wide. At a point on the road between the pillars, the angles of elevation of their tops are 60° and 30°. Find the height of the pillars.

Detailed Solution: Question 6

Let AB and CD be the two pillars of equal height 'h' m.
If P is the point between them, then
∠APB = 60° and ∠CPD = 30°

Let BP = x m
⇒ PD = (150 - x) m
In ΔABP, tan 60° = 
⇒ h = √3x … (1)

⇒ 3x = 150 - x or 4x = 150
∴ x = 37.5 and h = √3x = (1.732)37.5 = 64.95 m

Test Level 1: Trigonometry - 1 - Question 7
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If cosθ = 2cos2θ , then 2sec2θ – cos2θ + tan2θ =

Detailed Solution: Question 7

cosθ = 2 cos2θ

secθ = 2
secθ = sec 60°
θ = 60°
2 sec2θ – cos2θ + tan2θ = 2 sec2 60° - cos2 60° + tan2 60° = 

Test Level 1: Trigonometry - 1 - Question 8
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Find the value of

Detailed Solution: Question 8



(∵ sin (90° - θ) = cos θ and cos (90° - θ) = sin θ)

Test Level 1: Trigonometry - 1 - Question 9
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The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of its image in lake is 60°. The height of the cloud is

Detailed Solution: Question 9

Let A be the point of observation and C be the position of the cloud. Let DE be the surface of the lake and C' be the reflection of C in the lake.
∠CAB = 30°
And ∠C'AB = 60°
Let BC = h m

In right-angled ΔABC, BC/AB = tan 30°


In right-angled ΔABC',

h = 60 m
Height of cloud = 120 m

Test Level 1: Trigonometry - 1 - Question 10
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Solve for the number of solution of the equation: sin(x) + cos(x) = 1 on the interval 0° ≤ x < 360°.  

Detailed Solution: Question 10

sin(x) + cos(x) = 1
Squaring both side,
(sin(x) + cos(x))2 = 12
[sin2(x) + cos2(x)] + 2sin(x)cos(x) = 1 (Using identity: (a + b)2 = a2 + b2 + 2ab)
1 + 2sin(x)cos(x) = 1
2sin(x)cos(x) = 0
sin(x)cos(x) = 0
From the last line above, either sine is zero or cosine is zero, so solutions are:
x = 0°, 90°, 180°, 270°
Checking the value of 'x' in the above equation,
x = 0°, 90° satisfies.
x = 180° and 270° does not satisfies.
Hence, total number of solution = 2

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About this CAT Practice Test

This test contains 10 MCQs on Test Level 1: Trigonometry - 1 with detailed solutions for each question. It is part of the Level-wise Tests for CAT course on EduRev, designed to align with the current CAT syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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