Basic Trigonometry Concepts for CAT with Questions

Thursday, July 12th, 2018

Basic Trigonometry Concepts for CAT with Questions

Trigonometry is a broad topic and hence we will cover all the concepts and sample examples in a two part post. In this first part, we would discuss concepts, formulae and some sample examples. In the second part of the post, we will follow up with some more examples.

Trigonometry is the branch of mathematics that deals with the study of relationships between sides and angles of a triangle. It is derived from Greek word, Tri meaning three and Gon means Angle and Metron means Measure.

The ratio of the lengths of two sides of a right angled triangle is called a trigonometric ratio.

There are six trigonometric ratios.

Consider a right angled triangle ABC as shown below with:

AC= Hypotenuse of triangle

AB= Side adjacent to angle A

BC = Side opposite to angle A


Trigonometric ratios of right angled triangle ABC with ∠B = 90° :

  • Sin A = sine of ∠A = (side opposite to ∠A) / hypotenuse = (BC/AC)
  • cos A = cosine of ∠A = (side adjacent to ∠A) / hypotenuse = (AB/AC)
  • tan A = tangent of ∠A = (side opposite to ∠A) / (side adjacent to ∠A) = (BC/AB)
  • cosec A = cosecant of ∠A = 1/ sin A
  • sec A = secant of ∠A = 1/ cos A
  • cot AA = cotangent of ∠A = 1/ tan A

Trigonometric ratios of specific angles:

The trigonometric ratios of some special angles i.e. 0°, 30°,45°,60°,90° follow a pattern and are easy to remember. Identifying and remembering these patterns helps in solving problems involving these angles.

Two angles are said to be complementary if their sum is 90°. Thus ϴ and (90 – ϴ) are complementary angles. Trigonometric ratios of complementary angles help in simplifying the problems. Representing complementary angles in terms of these standard angles helps in solving a complex problem involving trigonometric ratios.

Table for trigonometric ratios of specific angles:


  • Trigonometric ratios of complementary angles:
    • Sin (90 – A) = cos A
    • Cos (90 – A) = sin A
    • Tan (90 – A) = cot A
    • Cot (90 – A) = tan A
    • Sec (90 – A) = cosec A
    • Cosec (90 – A) = sec A
  • Trigonometric Identities
    • sin2Ꝋ + cos2Ꝋ = 1
    • tan2Ꝋ + 1 = sec2
    • cot2Ꝋ + 1 = cosec2
  • Two Special triangles

There are two special triangles we need to know, 45-45-90 and 30-60-90 triangles. They are depicted in the figures below.


The figures show how to find the side lengths of those types of these special triangles.

  • In a 45-45-90 triangle ABC shown above, ratio of side AC:BC:AB = 1:1:1/√2
  • In a 60-30-90 triangle ABC shown above, ratio of sides AC:BC:AB = 1: 1/√3: 2

Let us look at some examples to understand the concept better.

Example 1: Anil looked up at the top of a lighthouse from his boat and found the angle of elevation to be 30 degrees. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45 degrees. Find the height of the lighthouse.

a) 25

b) 25√3

c) 25(√3-1)

d) 25(√3+1)



If we look at the above image, A is the previous position of the boat. Angle of elevation from this point to the top of the light house is 30 degrees.

After sailing for 50 m, Anil reaches point D from where angle of elevation is 45 degrees. C is the top of the light house.

Let BD = x

Now, we know tan 30 degrees = 1/ √3 = BC/AB

Tan 45 degrees = 1

=> BC = BD = x

Thus, 1/ √3 = BC/AB = BC / (AD+DB) = x / (50 + x)

Thus x (3 -1) = 50 or x= 25(3 +1) m.

Example 2: An airplane flying at 3000 m above the ground passes vertically above another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 600 and 450 respectively. The height of the lower plane from the ground is

a) 1000√3 m

b) 1000/√3 m

c) 500 m

d) 1500(√3+1)



Let the higher plane fly such that at point P the angle of elevation from point Q is 60o.

Let the height of the plane flying at a lower level be 'h' QR=h (since tan 45o= 1)

Tan 60 degrees = 3000/h

=> √3 = 3000/h

=> h =  3000 / √3 = 1000 √3

Example 3: Ram and Shyam are 10 km apart. They both see a hot air balloon passing in the sky making an angle of 60° and 30° respectively. What is the height at which the balloon could be flying?

a.) 5/ 2√3

b.) 5√3

c.) Both A and B

d.) Can't be determined

Solution: There can be two approaches to solve this problem. In one approach, Ram and Shyam could be on the opposite side of the balloon and in another approach they could be on the same side of the balloon.

Approach 1:  Ram and Shyam are on opposite sides of the balloon


tan 30° = h/x

∴ h = x/√3

tan 60° = h/(10−x)

=> h = (10 – x) √3

=> (10 – x) √3 = x/√3

=> 30 – 3x = x

=> x = 30/4=15/2

=> h = 15/(2√3)=(5√3)/2

 Approach 2: Ram and Shyam are on same side of balloon


tan 60° = h/x

=> h = x√3


tan 30° = h/(10−x)

Thus, h = (10−x)/√3

=> x√3 = (10−x)/√3

=> 3x – x = 1=

=> X = 5

=> h = 5√3

Thus, both A and B are correct choices.

Example 4: Consider a regular hexagon ABCDEF. There are towers placed at B and D. The angle of elevation from A to the tower at B is 30 degrees, and to the top of the tower at D is 45 degrees. What is the ratio of the heights of towers at B and D?



Let the hexagon ABCDEF be of side 'a'. Line AD = 2a. Let towers at B and D be B'B and D'D respectively.
From the given data we know that ∠B´AB = 30° and ∠D´AB = 45°. We might need to keep in mind that the Towers B'B and D´D are not in the same plane as the hexagon.

Now, in triangle B'AB,

Tan∠B´AB = B′B/AB=1/ √3
=> B'B = a / √3

In Triangle D´AD, tan ∠D´AD = D′D / AD = tan 45 = 1
=> D'D = 2a = AD

Ratio of heights = 1/√3: 2 or 1 / 2√3

Example 5:  Find the maximum and minimum value of 8 cos A + 15 sin A + 15

Solution:  In these types of questions, we should always look for Pythagorean triplets:

We know that (8, 15, and 17) is one of the Pythagorean triplets.

Writing the above equation in terms of Pythagorean triplets,

17 (8/17cosA+15/17sinA)  + 15

Let there be an angle B for which sinB = 8/17, cosB=15/17

This will imply that cos A = 8/17 and sin A = 15/17

From above equation, we have to now find a maximum and minimum value of

17(sin B cos A + cos B sin A) + 15

=> 17(sin(A+B)) + 15

We know that sin (A+B) max = 1
sin (A+B) min = -1
∴ Max value = 17×1 +15 = 32
Min value = 17 x -1 + 15 = -2
Answer: Maximum value is 32 and minimum value is -2

Conclusion: As we have seen, concepts of trigonometry can be asked in the form of maximum/minimum value or heights and distances. We have covered some sample CAT questions in this post and we will follow up with some more examples in the second part. Happy studying.

You can also see: Should You Quit Your Job To Prepare For CAT?

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Basic Trigonometry Concepts for CAT with Questions

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