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# Basic Trigonometry Concepts Quant Notes | EduRev

## GMAT : Basic Trigonometry Concepts Quant Notes | EduRev

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

The trigonometric ratios of a triangle are also called the trigonometric functions. There are six trigonometric ratios. Sine, cosine, and tangent are 3 important trigonometric functions and are abbreviated as sin, cos and tan.

• Trigonometry is the branch of mathematics that deals with the study of relationships between sides and angles of a triangle.
• It is derived from the Greek word, Tri meaning three and Gon means Angle and Metron means Measure. • Consider a right-angled triangle ABC, where the longest side is called the hypotenuse, and the sides opposite to the hypotenuse are referred to as the adjacent and opposite sides.
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°
(i) sin A = Sine of ∠A = (Side opposite to ∠A / hypotenuse) = (BC/AC)
(ii) cos A = Cosine of ∠A = (Side adjacent to ∠A / hypotenuse) = (AB/AC)
(iii) tan A = Tangent of ∠A = (Side opposite to ∠A / Side adjacent to ∠A) = (BC/AB)
(iv) cosec A = Cosecant of ∠A = Hypotenuse / Side opposite to ∠A = 1/sin A
(v) sec A = Secant of ∠A = Hypotenuse / Adjacent side = 1/cos A
(vi) cot A = Cotangent of ∠A = Adjacent side / Opposite side = 1/tan A

• The trigonometry angles which are commonly used in trigonometry problems are  0°, 30°, 45°, 60° and 90°.
• The trigonometric ratios such as sine, cosine and tangent of these angles are easy to memorize.
Example: In a right-angled triangle:
⇒ Sin θ = Perpendicular/Hypotenuse
or θ = sin-1 (P/H)
⇒ Similarly,
θ = cos-1 (Base/Hypotenuse)
θ = tan-1 (Perpendicular/Base)

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.
Table: Trigonometric Ratios of Specific Angles • Two angles are said to be complementary if their sum is 90°. Thus ϴ and (90 – ϴ) are complementary angles.
• Representing complementary angles in terms of these standard angles helps in solving a complex problem involving trigonometric ratios.

Try yourself: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
Basic Trigonometric Identities
• sin2θ + cos2θ = 1
• tan2θ + 1 = sec2θ
• cot2θ + 1 = cosec2θ
• Sinθ /Cosθ = tanθ
• Cosθ / Sinθ = cotθ
• sin 2θ = 2 sin θ cos θ

• cos 2θ = cos²θ – sin²θ

• tan 2θ = 2 tan θ / (1 – tan²θ)

• cot 2θ = (cot²θ – 1) / 2 cot θ

Try yourself:The value of sin θ and cos (90° – θ)

Sum and Difference Identities

For angles u and v, we have the following relationships:

• sin(u + v) = sin(u)cos(v) + cos(u)sin(v)
• cos(u + v) = cos(u)cos(v) – sin(u)sin(v)
• tan(u+v) = (tan(u) + tan(v)) / (1−tan(u) tan(v))
• sin(u – v) = sin(u)cos(v) – cos(u)sin(v)
• cos(u – v) = cos(u)cos(v) + sin(u)sin(v)
• tan(u-v) = (tan(u) − tan(v)) / (1+tan(u) tan(v))
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:

Try yourself: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 30º. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45º. 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. The angle of elevation from this point to the top of the lighthouse is 30 degrees. After sailing for 50 m, Anil reaches point D from where the angle of elevation is 45 degrees. C is the top of the lighthouse.

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.

Try yourself: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 the 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 30º and 60º, respectively. Find the distance between the two planes at that instant. 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 60º and 45º 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 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 60º = 3000/y
⇨ √3 = 3000/y
⇨ y =  3000/√3 = 1000√3

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## Quantitative Aptitude (Quant)

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