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# Basic Trigonometry Concepts Quant Notes | EduRev

## SSC : Basic Trigonometry Concepts Quant Notes | EduRev

The document Basic Trigonometry Concepts Quant Notes | EduRev is a part of the SSC Course Quantitative Aptitude (Quant).
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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. Trigonometry

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: Question 1:Value of tan 30°/cot 60° is:
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

• sin2θ + cos2θ = 1
• tan2θ + 1 = sec2θ
• cot2θ + 1 = cosec2θ
Question 2:The value of sin θ and cos (90° – θ)

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.

Question 3:The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

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)

Solution: 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/A
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.

Question 4:From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Example 2: An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 300 and 600, respectively. Find the distance between the two planes at that instant.
Solution: Let C and D be the two aeroplanes and A be the point of observation. Then,  Therefore, the distance between the two planes is 6250 m.

Example 3: 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)
Solution: Let the higher plane fly such that at point A the angle of elevation from point D is 60o.
Let the height of the plane flying at a lower level be CD = y (since tan 45o= 1)
Tan 60 degrees = 3000/y
⇒ √3 = 3000/y
⇒ y =  3000/√3 = 1000 √3

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## Quantitative Aptitude (Quant)

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