# Class 10 Trigonometry

Trigonometry is the branch of mathematics that deals with the sides and angles of a right-angled triangle. It is derived from the Greek words ‘tri’ which means three, ‘gon’ which means sides, ‘metron’ which means measure. It was used by early astronomers and in Egypt and Babylon. It is used today in various subjects like Architecture, Engineering, Physical Science. It is in class 10 that students have an introduction to Trigonometry. In this blog, we will study class 10 trigonometry and trigonometric ratios for different angles of measure ( 0 to 90 degrees) in detail.

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## Trigonometric Ratios

The first part in class 10 trigonometry is Trigonometric Ratios. Trigonometric ratios are the ratio of sides of a right-angle triangle and the acute angles.

There are 6 Trigonometric ratios. Sine, Cosine, Tangent, Cotangent, Secant, Cosecant.

In the right angle triangle given above B is 90 degrees.

We will define the Trigonometric Ratios of ∠A.

1. Sine of A = SinA = Side opposite to Angle A / Hypotenuse = BC / AC
2. Cosine of A = CosA = Side adjacent to Angle A / Hypotenuse = AB / AC
3. Tangent of A = TanA = Side opposite to Angle A / Side adjacent to Angle A = BC / AB
4. Cotangent of A = CotA= 1/TanA = Side Adjacent to Angle A / Side opposite to Angle A = AB/BC
5. Cosecant of A = CosecA = 1 / SinA = Hypotenuse / Side opposite to Angle A = AC/BC
6. Secant of A = SecA = 1/CosA = Hypotenuse/Side adjacent to Angle A = AC/AB

Observe that TanA is SinA / CosA and CosecA, SecA, CotA are reciprocals of SinA, CosA, TanA respectively.

It is important to understand class 10 trigonometry through examples.

Example 1: In triangle ABC, ∠B = 90 degrees, tan A = 6/5, Find other trigonometric ratios of Angle A

Solution
Given: ∠B = 90. So AC is hypotenuse
TanA = BC/AB =6/5 = 6k/5k
for k having positive value.

So BC= 6k and AB = 5k
Using Pythagoras  Theorem
AB2 + BC2 = AC
(5k)2 + (6k)2 = 25k2 + 36k2
= 61k2  = AC2

Square rooting both sides
61k = AC
Now we have a length of all sides.

1. SinA = BC/AC = 6k/61k = 6/61
2. CosA = AB/AC = 5k/61k = 5/61
3. TanA = 6/5 (GIVEN)
4. CotA = 1/TanA = 5/6
5. CosecA = 1/SinA = 61/6
6. SecA = 1/COsA = 61/5

Example 2:

In triangle ABC, right-angled at C, AB= 5 units and BC= 4 units, and ∠B = θ. Find the value of

1. Cos2θ + Sin2θ
2. Cos2θ – Sin2.θ

Solutions:

Using pythagoras theorem, AC2 = AB2 – BC2
AC2 .= 52 – 42 = ( 5-4) ( 5 + 4 )
AC2 = (1) (9) = 9
AC = 3 units
Sinθ = AC/AB  = 3/5 , Cosθ = BC/AB = 4/5

Now

1. Cos2θ + Sin2θ  = (4/5)2 + (3/5)2  = 16+9/25 = 25/25 = 1
2. Cos2θ – Sin2θ  = (4/5)2 – (3/5)2  = 16 – 9/25 = 7/25

Also Read: Statistics Class 10 Study Notes

## Trigonometric ratios of Some Specific Angles

Using the above information let’s solve the following questions as it is easier to understand class 10 trigonometry:

Q1. In an triangle ABC, right-angled at B, AB = 7 and ∠ACB = 30 degrees. Find the lengths of sides BC and AC.

Solution
Given: ∠ACB = 30 degrees, AB=7 cm,
Let BC = x cm and AC = y cm
TanC = AB/BC = 7/x
TanC = Tan30 = 1/√3
7/x = 1/√3
x = 7√3
BC = x = 7√3

To find AC
Sin30 = AB/AC
1/2 = 7/y
AC = y = 14

Alternatively, we could have used the Pythagoras theorem to find AC

I.e. AC2 = AB2 + BC2 = 72 + (7√3)2
= 49 + 147
= 196
AC= 14 cm

Q2. If tan(A+B)= √3 and tan(A–B)= 1/√3 ;0°<A+B≤90° ; A>B, find A and B.

Solution
tan(A+B) = √3
Since tan60° = √3
tan(A+B) = tan60°
Therefore A+B = 60   –(i)
tan(A-B) = 1/√3
Since tan30° = 1/√3
tan(A-B) = tan30°
Therefore A-B = 30   –(ii)
A + B + A – B = 60 + 30
2A = 90
A = 45

Substitute this value of A in (i)
A + B = 60
45 + B = 60
B = 15
Therefore A = 45°, B = 15°

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## Trigonometric Ratios of Complementary Angles

We move on next to trigonometric ratios of complementary angles in class 10 trigonometry.

If θ is an acute angle, its complementary angle is 90-θ. The trigonometric ratios of complementary angles.

1. Sin( 90-θ) = Cosθ
2. Cos (90-θ) = Sinθ
3. Tan(90-θ) = Cotθ
4. Cot(90-θ) = Tanθ
5. Cosec(90-θ) = Secθ
6. Sec(90-θ) = Cosecθ

Let us understand this concept of class 10 trigonometry through these practice questions:

Q.1 Evaluate  Cos55 / Sin35

Solution: Sin(35) = Cos (90 – 35) = Cos(55)

Cos(55)/Cos(55) = 1

Q.2 If Sin 3A = Cos (A-30), where 3A is an acute angle, Find the value of A.

Solution: Given Sin3A = Cos(A-30)

Since Sin3A = Cos(90-3A)

So Cos(90-3A) = Cos(A-30)

90-3A = A – 30

120 = 4A

A = 30

Q3. Express Tan75 + Cos65 in terms of trigonometric ratios of angles between 0 and 45

Solution: Tan75 + Cos65 = Tan(90-15) + Cos(90-25)

= Cot15 + Sin25

## Trigonometric Identities

We further move on to trigonometric identities in class 10 trigonometry. An equation involving trigonometric ratios of an angle is called a trigonometric identity.

1. Cos2A + Sin2A = 1
2. 1 + Tan2A = Sec2A
3. Cot2A + 1 = Cosec2

We can understand this section of class 10 trigonometry better with these practice questions:

Q.1 Prove that CosecA (1-CosA) (CosecA + CotA) = 1

Solution:

CosecA (1-CosA) (CosecA + CotA) = (1/SinA) (1-CosA) ( 1/SinA + CosA/SinA)

(1-CosA)(1+CosA) / Sin2A = 1-Cos2A / Sin2A

Sin2A / Sin2A = 1

Q.2 Prove that Cosθ − Sinθ + 1 / Cosθ + Sinθ -1 = 1 / Cosecθ – Cotθ,

using the identity Sec2θ = 1 + Tan2θ

Solution: LHS = Cosθ – Sinθ + 1 / Cosθ + Sinθ – 1

Divide both LHS and RHS by Sinθ

= Cotθ – 1 + Cosecθ / Cotθ + 1 – Cosecθ

= (Cotθ + Cosecθ) – 1 / (Cotθ – Cosecθ) + 1

Multiplying both numerator and denominator by Cotθ – Cosecθ

= {(Cotθ + Cosecθ) – 1 } (Cotθ – Cosecθ) / {(Cotθ – Cosecθ) + 1} (Cotθ – Cosecθ)

= ( Cot2θ – Cosec2θ) – (Cotθ – Cosecθ) / (Cotθ – Cosecθ) + 1) (Cotθ – Cosecθ)

Now, Cot2θ + 1 = Cosec2θ

Cot2θ – Cosec2θ = -1

= -1 -Cotθ + Cosecθ / (Cotθ – Cosecθ) + 1) (Cotθ – Cosecθ)

= -1 (Cotθ – Cosecθ + 1 )  / (Cotθ – Cosecθ + 1) (Cotθ – Cosecθ)

=  -1 / Cotθ – Cosecθ

=   1 / Cosecθ – Cotθ

Which is the RHS of the identity.

So Hence Proved.

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Thus, we come to the end of our blog. We hope this blog helped you understand and solve trigonometric questions of class 10 trigonometry. For more admission, related queries contact our experts at Leverage Edu. Sign up for a free session today!