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# NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Exercise 8.1 of Chapter 8 Introduction to Trigonometry are created here for your help in boosting your preparation of the class 10 maths CBSE board exam. All questions of class 10 NCERT maths exercise 8.1 Introduction to Trigonometry are solved by a step-by-step method so every student of class 10 can understand the solutions with the clearance of doubts. Here you can study NCERT solutions of other chapters also. These NCERT Solutions class 10 maths exercise 8.1 will clear all your doubts about basic Trigonometry which is required to solve the questions of higher class maths.

## NCERT Solutions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Exercise 8.1- Introduction to Trigonometry

Exercise 8.2 -Introduction to Trigonometry

Exercise 8.3-Introduction to Trigonometry

Exercise 8.4-Introduction to Trigonometry

NCERT solutions of Important Questions-Introduction to Trigonometry

## NCERT Solutions for Class 10 Maths Exercise 8.1 of Chapter 8 Introduction to Trigonometry

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Q1. In ΔABC right angled at B,AB = 24 cm,BC = 7 cm. Determine

(i) sin A, cos A

(ii) sin C, cos C

Ans. The given triangle is ΔABC in which AB = 24 cm, BC = 7 cm and ∠B =90°

Applying Pythagoras theorem

AC = √(AB² + BC²) =√(24² + 7²)=√(576 +49) =√(625) = 25 cm

(i)

$\fn_cm sinA =\frac{p}{h}=\frac{BC}{AC}=\frac{7}{25}$

$\fn_cm cosA =\frac{b}{h}=\frac{AB}{AC}=\frac{24}{25}$

(ii)

$\fn_cm sinC =\frac{P}{h}=\frac{AB}{AC}=\frac{24}{25}$

$\fn_cm cosC =\frac{b}{h}=\frac{BC}{AC}=\frac{7}{25}$

Q2. In the given figure, find tan P – cot R

Ans.In the given ΔPQR, we have PR = 13 cm, PQ =12 cm

Applying Pythagoras theorem

QR = √(PR² – PQ²) = √(13² – 12²) = √(169 -144) = √25 = 5

$\fn_cm tanP =\frac{P}{h}=\frac{QR}{PQ}=\frac{5}{12}$

$\fn_cm cotR =\frac{b}{h}=\frac{QR}{PQ}=\frac{5}{12}$

The value of tan P – cot R

$\fn_cm =\frac{5}{12}-\frac{5}{12}$

tan P – cot R= 0

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

Ans. We are given sin A = 3/4

Since sin A = Perpendicular/Hypotenuse

Let perpendicular = 3x and hypotenuse = 4x

Base = √[(4x)²- (3x)²] = √(16x²-9x²) =x√7

$\fn_cm cosA=\frac{b}{h}=\frac{x\sqrt{7}}{4x}=\frac{\sqrt{7}}{4}$

$\fn_cm tanA=\frac{p}{b}=\frac{3x}{x\sqrt{7}}=\frac{3}{\sqrt{7}}$

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

Ans. We are given that

15 cot A = 8

$\fn_cm cotA=\frac{8}{15}$

$\fn_cm \because cotA=\frac{Base}{Perpendicular}$

Let Base is 8x and perpendicular is 15x

Applying pythogorus theorem

Hypotenuse = √(Base² + Perpendicular²) = √[(8x)² + (15x)²]= √(64x² + 225x²)= √(289x²) =17x

$\fn_cm sinA=\frac{p}{h}=\frac{15x}{17x}=\frac{15}{17}$

$\fn_cm secA=\frac{h}{b}=\frac{17x}{8x}=\frac{17}{8}$

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

Ans. We are given that

$\fn_cm \because sec\Theta =\frac{hypotenuse}{base}$

Let hypotenuse is 13x and base is 12x

Applying Pythagoras theorem

Perpendicular = √(hypotenuse² – base²) =√(13² – 12²)=√(169-144)=√25 = 5

$\fn_cm sin\Theta =\frac{p}{h}=\frac{5}{13}$

$\fn_cm cos\Theta =\frac{b}{h}=\frac{12}{13}$

$\fn_cm tan\Theta =\frac{p}{b}=\frac{5}{12}$

$\fn_cm cot\Theta =\frac{b}{p}=\frac{12}{5}$

$\fn_cm sec\Theta =\frac{b}{h}=\frac{12}{13}$

$\fn_cm cosec\Theta =\frac{h}{p}=\frac{13}{5}$

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

Ans. We are given that

cos A = cos B where ∠A and ∠B are acute angles, the ΔABC must be as follows

$\fn_cm cosA=\frac{b}{h}=\frac{AC}{AB},cosB=\frac{b}{h}=\frac{BC}{AB}$

Since

cos A = cos B

$\fn_cm \frac{AC}{AB}=\frac{BC}{AB}$

AC = BC

∠A = ∠B (angles opposite to equal sides in a triangle)

You can see the video for Q1 to Q6 of exercise 8.1

Q7. If cot θ = 7/8,evaluate

$\fn_cm \left ( i \right )\frac{\left ( 1+sin\Theta \right )\left ( 1-sin\Theta \right )}{\left ( 1+cos\Theta \right )\left ( 1-cos\Theta \right )}$

(ii) cot²θ

Ans. (i) We are given cot θ = 7/8

$\fn_cm cot\Theta =\frac{Base}{Perpendicular}$

Let the base = 7x and perpendicular = 8x

Applying pythogorus theorem

Hypotenuse = √(Base² + Perpendicular²) = √[(7x)² + (8x)²]= √(49x² + 64x²)= √(113x²) =x√113

$\fn_cm sin\Theta =\frac{p}{h}=\frac{8x}{x\sqrt{113}}=\frac{8}{\sqrt{113}}$

$\fn_cm cos\Theta =\frac{b}{h}=\frac{7x}{x\sqrt{113}}=\frac{7}{\sqrt{113}}$

The given expression

$\fn_cm \frac{\left (1+sin\Theta \right )\left ( 1-sin\Theta \right )}{\left ( 1+cos\Theta \right )\left ( 1-cos\Theta \right )}$

Putting the value of sin θ and cos θ in the given expression

$\fn_cm =\frac{\left (1+8/\sqrt{113} \right )\left ( 1-8/\sqrt{113} \right )}{\left ( 1+7/\sqrt{113} \right )\left ( 1-7/\sqrt{113} \right )}$

Applying the identity ( x + y)(x -y) = x ² – y²

$\fn_cm =\frac{ 1^{2}-(8/\sqrt{113} )^{2}}{ 1^{2}-(7/\sqrt{113} )^{2}}$

$\fn_cm =\frac{ 1-64/113}{ 1-49/113}$

$\fn_cm =\frac{ 49}{ 64}$

(ii) The given expression is

cot²θ

$\fn_cm =\frac{cos^{2}\Theta }{sin^{2}\Theta }$

Putting the value of cos θ and sin θ

$\fn_cm cot^{2}\Theta =\frac{cos^{2}\Theta }{sin^{2}\Theta }=\frac{\left ( 7/\sqrt{113} \right )^{2}}{\left ( 8/\sqrt{113} \right )^{2}}=\frac{49}{64}$

Q8.If 3 cot A = 4, check whether ( 1-tan²A)/(1 +tan²A) = cos²A – sin²A or not

Ans. We are given

3 cot A = 4

$\fn_cm cotA=\frac{4}{3}$

$\fn_cm cotA=\frac{Base}{Perpendicular}$

Let base = 4x and perpendicular =3x

Applying pythogorus theorem

Hypotenuse = √(Base² + Perpendicular²) = √[(4x)² + (3x)²]= √(16x² + 9x²)= √(25x²) =5x

The value of tan A = p/b = 3x/4x = 3/4, cos A =b/h =4x/5x = 4/5,sin A =p/h=3x/5x = 3/5

Putting the value of tan x, cosx and of sinx in the following equation

$\fn_cm \frac{1-tan^{2}A}{1+tan^{2}A}= cos^{2}A-sin^{2}A$

$\fn_cm \frac{1-\left ( 3/4 \right )^{2}}{1+\left (3/4 \right )^{2}}= \left ( \frac{4}{5} \right )^{2}-\left ( \frac{3}{5} \right )^{2}$

$\fn_cm \frac{1-9/16}{1+9/16}= \frac{16}{25}-\frac{9}{25}$

$\fn_cm \frac{7/16}{25/16}= \frac{16}{25}-\frac{9}{25}$

$\frac{7}{25}=\frac{7}{25}$

LHS = RHS, The equation shown is OK

See the video for Q7 and Q8

Q9. In ΔABC ,right-angled at B ,if tan A = 1/√3 ,find the value of :

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Ans. We are given that tan A = 1/√3

$\fn_cm tanA=\frac{Perpendicular}{Base}=\frac{BC}{AB}$

Let BC =x and AB =√3 x

Applying Pythagoras theorem

AC = √(AB² + BC²) = √[(√3x)² + (x)²]= √(3x² + x²)= √(4x²) =2x

(i) sin A cos C + cos A sin C

$\fn_cm \frac{BC}{AC}.\frac{BC}{AC}+\frac{AB}{AC}.\frac{AB}{AC}$

$\fn_cm \frac{BC^{2}}{AC^{2}}+\frac{AB^{2}}{AC^{2}}$

$\fn_cm \frac{x^{2}}{\left ( 2x \right )^{2}}+\frac{\left ( \sqrt{3} x\right )^{2}}{\left ( 2x \right )^{2}}$

$\fn_cm \frac{1}{4}+\frac{3}{4}=1$

(ii) cos A cos C – sin A sin C

$\fn_cm \frac{AB}{AC}.\frac{BC}{AC}-\frac{BC}{AC}.\frac{AB}{AC}$

$\fn_cm \frac{\sqrt{3}x}{2x}.\frac{x}{2x}-\frac{x}{2x}.\frac{\sqrt{3}x}{2x}$

$\fn_cm \frac{\sqrt{3}}{2}.\frac{1}{2}-\frac{1}{2}.\frac{\sqrt{3}}{2}$

$\fn_cm \frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$

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

Ans. The given ΔPQR,in which ∠Q=90°, PR + QR = 25 cm and PQ = 5 cm

PR + QR = 25 cm

QR = 25 – PR

Applying Pythogorus theorem

PR² = QR² + PQ²

PR² = (25 – PR)² + PQ²

PR² = 625 + PR²- 50PR + 5²

50PR =650

PR = 13 cm, QR = 25 -13 = 12 cm

sin P = p/h = QR/PR = 12/13

cos P = b/h = PQ/PR = 5/13

tan P = p/h = QR/PQ = 12/5

See the video for Q9,Q10 and Q11

## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

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### NCERT solutions of class 11 maths

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CBSE Class 11-Question paper of maths 2015

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### NCERT solutions of class 12 maths

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