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Test Level 2: Trigonometry - 2 - CAT MCQ


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

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

If cosθ = 5/21 sinθ, then the value of 

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

Gievn: cosθ = 5/21 sinθ

Dividing both numerator and denominator by sinθ, 

Test Level 2: Trigonometry - 2 - Question 2

The length of the shadow of a vertical pole on level ground increases by 25 metres when the altitude of the sun changes from 60° to 30°. Calculate the height of the pole.

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

Let AB be the vertical pole and AD be the shadow when sun's altitude is 60°.

∴ ∠BDA = 60°
Similarly, ∠BCA = 30°
CD = 25 m (Given)
In ΔADB,
tan 60° = AB

In ΔBDE,

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Test Level 2: Trigonometry - 2 - Question 3

The angles of elevation of a plane flying at a constant altitude of 10000 ft are found to be 60° and 30° at an interval of 1 minute. What is the speed of the plane?

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


In the figure, BE = CD = 10000 ft
∠BAE = 60° and ∠CAD = 30°
In triangle ABE,

In triangle ACD,

Distance travelled in one minute = ED = BC = AC – AB

Test Level 2: Trigonometry - 2 - Question 4

If 4 cos2 A + 2 sin2 A = 3, then what is the value of A, given that A lies in the first quadrant?

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

4 cos2 A + 2 (1 - cos2 A) = 3
4 cos2 A + 2 - 2cos2A = 3
2 + 2 cos2A = 3
2(1 + cos2 A) = 3
1 + cos2 A = 3/2

A = 45°

Test Level 2: Trigonometry - 2 - Question 5

AB is a vertical pole. The end A is on the level ground and C is the middle point of AB. P is a point on the level ground. The portion BC subtends an angle β at P. If AP = nAB, then tan  equals

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

AC = AP tan α
⇒ (1/2) AB = n AB tan α

Test Level 2: Trigonometry - 2 - Question 6

A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p, such that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal.
What is the value of P/q?

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


Length of the ladder will remain the same.
∴ AC = ED

Now,

Test Level 2: Trigonometry - 2 - Question 7

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



[Using formula, cos C - cos D = 2 sin]

[since, sin(-θ) = - sinθ, cos (90° - 50°) = sin 50°]

[Using formula, sin C - sin D = 2 sin ]

= 2 cos 30° =√3

Test Level 2: Trigonometry - 2 - Question 8

If tan α + sin α = m and tan α - sin α = n, then m2 - n2 =

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

m2 - n2 = (m + n)(m - n) = 2 tan α × 2 sin α = 4 tan α sin α
Now, going by options:

Test Level 2: Trigonometry - 2 - Question 9

Find the value of 

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


Test Level 2: Trigonometry - 2 - Question 10


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


Test Level 2: Trigonometry - 2 - Question 11

Detailed Solution for Test Level 2: Trigonometry - 2 - Question 11



Test Level 2: Trigonometry - 2 - Question 12

A 1.6 m tall observer is 45 meters away from a tower. If the angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in meters is
(Take √3 = 1.732)

Detailed Solution for Test Level 2: Trigonometry - 2 - Question 12


Given height of the observer, DE = 1.6 m
Hence, BC = 1.6 m
AC is the height of the tower.
Let AB = h.
Given distance between the observer and the tower, CD = 45 m
⇒ BE = 45 m
In right ΔABE,

tan 30° = (h/45)

45 = h√3
h = 45 ÷ 1.732 = 25.98 m
Thus, height of tower = AC = AB + BC = 25.98 m + 1.6 m = 27.58 m
Hence, answer option 3 is correct.

Test Level 2: Trigonometry - 2 - Question 13

The angle of elevation of the top of a tower from a point on the ground at some distance from its base is 60°. The angle of elevation of the top of the tower from a point 20 m above the same point on the ground is 30°. What is the height of the tower?

Detailed Solution for Test Level 2: Trigonometry - 2 - Question 13


AD is the tower and BC = 20 m
In triangle ACD,

AD = AE + ED
ED = BC = 20 m

AE + 20 = √3 CD . ....,. (1)
In triangle ABE,

Since, BE = CD
⇒ CD = √3 AE
On putting the value in equation 1, we get
AE + 20 = 3AE
⇒ 2AE = 20
⇒ AE = 10 m
So, AD = 10 + 20 = 30 m
∴ The height of the tower is 30 m.

Test Level 2: Trigonometry - 2 - Question 14

What is the value of cosθ if n = sinθ - cosθ ?

Detailed Solution for Test Level 2: Trigonometry - 2 - Question 14

n = sinθ - cosθ

Test Level 2: Trigonometry - 2 - Question 15

Two vertical poles of equal height are 120 m apart. On the line joining their bases, A and B are two points. Angle of elevation of the top of one pole from A is 45o and that of the other pole from B is also 45o. If AB = 30 m, then the height of each pole is

Detailed Solution for Test Level 2: Trigonometry - 2 - Question 15



Hence, 120 = h + 30 + h
⇒ h = 45 m
∴ Height of the vertical poles = 45 m

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