Class 10 Mathematics

Trigonometry

Exercise 8.1 (NCERT) Part 1

Important Formulae

In the right angle triangle ABC, the right angle is ∠ B

10 trigonometry exericse solution

`text(sin A)=text(side opposite to ∠A)/text(hypotenuse)=p/h`

`text(cos A)=text(side adjacent to ∠A)/text(hypotenuse)=b/h`

`text(tan A)=text(side opposite to ∠A)/text(side adjacent to ∠A)=p/b`


If sin, cos or tan ratio are calculated for angle C, then values for p, b and h shall change accordingly.

`text(cosecθ)=(1)/text(sinθ)`

`text(secθ)=(1)/text(cosθ)`

`text(cotθ)=(1)/text(tanθ)`

`text(tanθ)=text(sinθ)/text(cosθ)`

∠ A00300450600900
Sin A0`1/2``1/sqrt2``sqrt3/2`1
Cos A1`sqrt3/2``1/sqrt2``1/2`0
Tan A0`1/sqrt3`1`sqrt3`Undefined
Cosec AUndefined2`sqrt2``2/sqrt3`1
Sec A1`2/sqrt3``sqrt2`2Undefined
Cot AUndefined`sqrt3`1`1/sqrt3`0


The value of sin or cos never exceeds 1, but the value of sec and cosec is always greater than or equal to 1.

Trigonometric Identities:

`text(cos)^2A + text(sin)^2A = 1`

`1 – text(sin)^2A = text(cos)^2A`

`1 – text(cos)^2A = text(sin)^2A`

`1 + text(tan)^2A = text(sec)^2A`

`text(sec)^2A – text(tan)^2A = 1`

`text(sec)^2A – 1 = text(tan)^2A`

`text(cot)^2A + 1 = text(cosec)^2A`

`text(cosec)^2A – text(cot)^2A = 1`

`text(cosec)^2A – 1 = text(cot)^2A`

Exercise 8.1 Part 1

Question – 1 - In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

10 trigonometry exericse solution

(a) Sin A, Cos A

Solution: AB = 24 cm, BC = 7 cm, AC = ?

The value of AC can be calculated by using Pythagoras Theorem:

`AC^2 = AB^2 + BC^2`

Or, `AC^2 = 24^2+ 7^2`

`= 576 + 49 = 625`

Or, `AC = sqrt(625)=25`

`text(sin A)=(BC)/(AC)=(7)/(25)`

`text(cos A)=(AB)/(AC)=(24)/(25)`

(b) Sin C, Cos C

Solution:

`text(sin C)=(AB)/(AC)=(24)/(25)`

`text(cos C)=(BC)/(AC)=(7)/(25)`

Question – 2 - In the given figure, find tan P – cot R.

10 trigonometry exericse solution

Solution: Value of QR can be calculated by using Pythagoras theorem:

`QR^2 = PR^2 – PQ^2`

Or, `OR^2 = 13^2 – 12^2`

`= 169 – 144 – 25`

Or, `OR = 5`

Now;

`text(tan P)-text(cot R)`

`=(QR)/(PQ)-(QR)/(PQ)=0`


Question – 3 - If sin A = ¾, calculate cos A and tan A.

Solution: Sin A = ¾ = p/h

We can calculate b by using Pythagoras theorem;

`b^2 = h^2 – p^2`

Or, `b^2 = 4^2 – 3^2`

`= 16 – 9 = 7`

Or, `b = sqrt7`

Now;

`text(cos A)=b/h=sqrt7/4`

`text(tan A)=p/b=3/sqrt7`

Question – 4 - Given 15 cot A = 8, find sin A and sec A.

Solution: 15 cot A = 8

10 trigonometry exericse solution

This means, b = 8 and p = 15

We can calculate h by using Pythagoras theorem;

`h^2 = p^2 + b^2`

Or, `h^2 = 15^2 + 8^2`

`= 225 + 64 = 289`

Or, `h = 17`

Now;

`text(sin A)=p/h=(15)/(17)`

`text(sec A)=h/b=(17)/(8)`

Question – 5 - Given `text(sec θ) = (13)/(12)`, calculate all other trigonometric ratios.

Solution:

`text(sec θ)=(13)/(12)=h/b`

This means, h = 13 and b = 12.

We can calculate p by using Pythagoras theorem;

`p^2 = h^2 – b^2`

Or, `p^2 = 13^2 – 12^2`

`= 169 – 144 = 25`

Or, `p = 5`

Other trigonometric ratios can be calculated as follows:

`text(sin θ)=p/h=(5)/(13)`

`text(cos θ)=b/h=(12)/(13)`

`text(tan θ)=p/b=(5)/(12)`

`text(cosec θ)=h/p=(13)/(5)`

`text(cot θ)=b/p=(12)/(5)`

Question – 6 - If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B

Solution: For no two different angles the cos ratio is same. (Ref: Table of trigonometric ratio).


Next

Exercise 8.1(part II)

Exercise 8.2

Exercise 8.3

Exercise 8.4(part I)

Exercise 8.4(Part II)

Application(part I)

Application(part II)