10th Maths 8.1

Chapter 8

Introduction to Trigonometry

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NCERT Class 10th solution of Exercise 8.2

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Exercise 8.1

Q1. In `triangle ABC`, right-angled at `B`, `AB=24 cm, BC=7 cm`. Determine:

i) sin `A`, cos `A`

ii) sin `C`, cos `C`.

Sol. :

In `triangle ABC`, `angle B = 90^circ`

By Pythagoras theorem

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

`AC^2 = (24)^2+(7)^2`

`AC^2 = 576 + 49`

`AC^2 = 625`

`AC^2 = (25)^2`

`AC = 25 cm`

i) 

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

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

Answer:

`sin A = 7/25, cos = 24/25`

ii)

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

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

Answer:

`sin C = 24/25, cos C = 7/25`

Q2. In figure, find `tanP-cotR.`

Sol. :

In `triangle PQR,`

By Pythagoras theorem

`QR^2 = PR^2 - PQ^2`

`QR^2 = (13)^2 - (12)^2`

`QR^2 = 169 - 144`

`QR^2 = 25`

`QR^2 = (5)^2`

`QR = 5 cm`

`= tan P - cot R`

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

`= 5/12 - 5/12`

`= 0`

Answer:

`tan P - cot R = 0.`  

Q3. If `sin A=3/4,` calculate `cos A` and `tan A.`

Sol. :

Given:

`sin A = 3/4`

To Find:

`cos A, tan A`

Solve:

In `triangle ABC, angle B = 90^circ`

`sin A = 3/4 = (BC)/(AC)`

By Pythagoras theorem

`AB^2 = AC^2 - BC^2`

`AB^2 = (4)^2 - (3)^2`

`AB^2 = 16 - 9`

`AB^2 = 7`

`AB = sqrt7`

`cos A = (AB)/(AC) = (sqrt7)/4`

`tan A = (BC)/(AB) = 3/(sqrt7)`

Answer:

`cos A = (sqrt7)/4, tan A = 3/(sqrt7)`  

Q4. Given `15 cot A=8,` find `sin A` and `sec A`.

Sol. :

Given:

`cot A = 8/15`

To find:

`sin A` and `sec A`.

Solve:

In `triangleABC`

`cot A = 8/15 = (AB)/(BC)`

By Pythagoras theorem

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

`AC^2 = (8)^2 + (15)^2`

`AC^2 = 64 + 225`

`AC^2 = 289`

`AC^2 = (17)^2`

`AC = 17`

`sin A = (BC)/(AC) = (15)/(17)`

`sec A = (AC)/(AB) = (17)/8`

Answer:

`sin A = (15)/(17)` and `sec A = (17)/8`. 

Q5. Given `sec theta=13/12,` calculate all other trigonometric ratios.

Sol. :

Given:

`sec theta = 13/12`

To Find:

All trigonometric ratios.

Solve:

In `triangle ABC`

`sec theta = 13/12 = (AC)/(BC)`

By Pythagoras theorem

`AB^2 = AC^2 - BC^2`

`AB^2 = (13)^2 - (12)^2`

`AB^2 = 169 -144`

`AB^2 = 25`

`AB^2 = (5)^2`

`AB = 5`

`sin theta = (AB)/(AC) = 5/12`

`cos theta = (BC)/(AC) = 12/13`

`tan theta = (AB)/(BC) = 5/12`

`cosec theta = (AC)/(AB) =13/5`

`cot theta = (BC)/(AB) = 12/5`

Answer:

`sin theta  = 5/12`

`cos theta  = 12/13`

`tan theta  = 5/12`

`cosec theta  =13/5`

`cot theta  = 12/5`

 Q6. If `angle A` and `angle B` are acute angles such that `cos A = cos B,` then show that 

`angle A=angle B.`

Sol. :

Given:

`angle A` and `angle B` are acute angles.

`cos A = cos B`

To Show:

`cos A = cos B`

Solve:

Let `triangle ABC`

`cos A = cos B`

`(AC)/(AB) = (BC)/(AB)`

`AC = BC`

`angle A = angle B` [Angle opposite to equal sides are equal.]

Proved.

Q7. If `cot theta=7/8,` evaluate: 

i) `((1+sin theta)(1-sin theta))/((1+cos theta)(1-cos theta))`

ii) `cot^2 theta`

Sol. :

i)

`= ((1 + sin theta)(1 - sin theta))/((1 + cos theta)(1 - cos theta))` [`(a+b)(a-b)=a^2-b^2`]

`= (1 - sin^2 theta)/(1 - cos^2 theta)`

`= (cos^2 theta)/(sin^2 theta)`

`= (cot theta)^2`

`= cot^2 theta`

`= (7/8)^2`

`=49/64`

Answer:

`=49/64`

ii)

`= cot^2 theta`

`= (cot theta)^2`

`= (7/8)^2`

`= 49/64`

Answer:

`= 49/64`

Q8. If `3cot A=4,` check whether `(1-tan^2 A)/(1+tan^2 A)=cos^2 A-sin^2 A` or not.

Sol. :

Given:

`cot A = 4/3`

Solve:

Let `triangle ABC`

`cot A = 4/3 = (AB)/(BC)`

By Pythagoras theorem

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

`AC^2 = (4)^2 + (3)^2`

`AC^2 = 16 + 9`

`AC^2 = 25`

`AC^2 = (5)^2`

`AC = 5`

To Prove:

`(1-tan^2 A)/(1+tan^2 A)=cos^2 A-sin^2 A` or not.

Proof:

`LHS.`

`= (1- (BC)/(AB))^2/(1+(BC)/(AB))^2`

`= (1-(3/4)^2)/(1+(3/4)^2)`

`= (1-9/16)/(1+9/16)`

`= (7/16)/(25/16)`

`= 7/25`

`RHS.`

`= cos^2A-sin^2A`

`= ((AB)/(AC))^2-((BC)/(AC))^2`

`= (4/5)^2-(3/5)^2`

`= 16/5-9/25`

`= (16-9)/25`

`= 7/25`

`LHS = RHS`

Proved.

Q9. In triangle `ABC,` right-angled at `B,` if `tan A=1/sqrt3,` find the value of

i) `sin A cos C + cos A sin C`

ii) `cos A cos C - sin A sin C`

Sol. :

Given:

`tan A = 1/sqrt3`

Solve:

In `triangle ABC`

`tan A = 1/sqrt3 = (BC)/(AB)`

By Pythagoras theorem

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

`AC^2 = (sqrt3)^2 + (1)^2`

`AC^2 = 3 + 1`

`AC^2 = 4`

`AC^2 = (2)^2`

`AC = 2`

i)

`= sin A cos C + cos A + sin C`

`= (BC)/(AC).(BC)/(AC) + (AB)/(AC).(AB)/(AC)`

`= ((BC)/(AC))^2 + ((AB)/(AC))^2`

`= (1/2)^2 + (sqrt3/2)^2`

`= 1/4 + 3/4`

`= 4/4 = 1`

Answer:

 The value is `1`.

ii)

`= cos A cos C - sin A sin C`

`= (AB)/(AC).(BC)/(AC) - (BC)/(AC).(AB)/(AC)`

`= (AB.BC)/(AC)^2 - (AB.BC)/(AC)^2`

`= (sqrt3.1)/(2)^2 - (sqrt3.1)/(2)^2`

`= (sqrt3)/4 - (sqrt3)/4`

`= 0`

Answer:

The value is `0`.

Q10. In `triangle PQR,` right-angled at `Q, PR+QR=25 cm` and `PQ=5 cm.` Determine the values of `sin P, cos P` and `tan P.`

Sol. :

Given:

`triangle PQR,` `angle Q = 90^circ,`

`PR + QR = 25 cm,`___________(1) 

`PQ = 5 cm`.

To find:

`sin P, cos P,` and `tan P`.

Solve:

In `triangle PQR`

By Pythagoras theorem

`(PQ)^2 = (PR)^2 - (QR)^2`

`(5)^2 = (PR + QR)(PR - QR)`

`25 = 25(PR  - QR)`

`25/25 = PR - QR`

`PR - QR = 1`____________(2)

From equation (1) `+` (2)

`PR + QR + PR - QR = 25 + 1`

`PR + PR = 26`

`2PR = 26`

`PR = 26/2`

`PR = 13`

Put in equation (1)

`13 + QR = 25`

`QR = 25 - 13`

`QR = 12`

`sin P = (QR)/(PR) = 12/13`

`cos P = (PQ)/(PR) = 5/13`

`tan P = (QR)/(QP) = 12/5`

Answer:

`sin P = 12/13, cos P = 5/13, tan P = 12/5`.

Q11. State whether the following are true or false, Justify your answer:

i) The value of `tan A` is always less than `1.`

ii) `sec A=12/5` for some value of angle `A.`

iii) `cos A` is the abbreviation used for the cosecant of angle `A.`

vi) `cot A` is the product of `cot` and `A.`

v) `sin theta=4/3` for some angle `theta.`

Answer:

i) false,    ii) true    iii) false    iv) false    v) false.