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NCERT Solutions Class 10 Maths Chapter 8: Introduction to Trigonometry



In this page we have NCERT book Solutions for Class 10th Maths: Chapter 8 - Introduction to Trigonometry for EXERCISE 5.1 . Hope you like them and do not forget to like , social_share and comment at the end of the page.

Question 1 

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

(i) Sin A, cos A

(ii) Sin C, cos C

Solution

In Δ ABC, right-angled at B ,using Pythagoras theorem we have

AC= AB2 +BC2  = 576 + 49 = 625

Or AC=25 ( taking positive value only)

Now

(i) In a right angle triangle ABC where B=90° ,

NCERT solutions triangle

Sin A = BC/AC = 7/25

CosA = AB/AC = 24/25

(ii)

 NCERT trigonometry solutions figure

Sin C = AB/AC = 24/25

Cos C = BC/AC =7/25

Question 2

In below find tan P – cot R

NCERT book trigonometry question

Solution

Again consider the above figure . Now by Pythagoras theorem

PQ2 + QR2  =PR2

QR=5

Now

tan P = Perp/Base = 5/12

Cot R = Base/Perm = 5/12

So tan P – cot R=0

3. If $sin A= \frac{3}{4}$. Calculate cos A and tan A.

Solution

Given $sin A= \frac{3}{4}$

Or P/H=3/4

Let  P=3k and H=4k

class 10 trigonometry solutions figure

Now By Pythagoras theorem

P2 + B2 =H2

9k2 + B2 =16k2

Or  $ B =+ k \sqrt 7 $

Now $CosA = {B \over H} = {{\sqrt 7 } \over 4}$

Now $\tan A = {{SinA} \over {CosA}} = {3 \over {\sqrt 7 }}$

Question 4

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

Solution 4

$Cot A =\frac{8}{15}$
Or
${B \over P} = {8 \over {15}}$
Let B=8K and P=15k
 

class 10 trigonometry solutions figure

So in a right angle triangle with angle A

P2 + B2 =H2

Or H=17K

Sin A = P/H = 15/17

Sec A = H/B = 17/8

Question 5

Given sec θ =13/12. Calculate all other trigonometric ratios.

Solution 5

Given sec θ=13/12

Or

H/B=13/12

let H=13K ,B=12K

Trigonometry NCERT solution question 5 figure

So in a right angle triangle with angle A

P2 + B2 =H2

P=5k

sin θ = P/H = 5/13

Cos θ = B/H = 12/13

Tan θ = P/B = 5/12

Cosec θ= 1/sin θ = 13/5

cot θ = 1/tan θ = 12/5

Question 6

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

Solution 6

In a triangle

Cos A =cos B

Trigonometry NCERT book solution question 9 figure

AC/AB= BC/AB

⇒AC=BC

⇒Angle A and Angle B

 

Question 7

If cot θ =7/8 evaluate

(i) ${{\left( {1 + sin\theta } \right)\left( {1 - sin\theta } \right)} \over {\left( {1 + cos\theta } \right)\left( {1 - cos\theta } \right)}}$

(ii) ${\cot ^2}\theta $

Solution

Given

Given cot θ=7/8

Or

B/P=7/8

let B=7K ,P=8K

Trigonometry NCERT solution question 7 figure

So in a right angle triangle with angle θ

P2 + B2 =H2

$H = k\sqrt {113} $

$\sin \theta  = {P \over H} = {8 \over {\sqrt {113} }}$

$\cos \theta  = {B \over H} = {7 \over {\sqrt {113} }}$

(i) 

${{\left( {1 + sin\theta } \right)\left( {1 - sin\theta } \right)} \over {\left( {1 + cos\theta } \right)\left( {1 - cos\theta } \right)}}$

$ = {{1 - si{n^2}\theta } \over {1 - co{s^2}\theta }}$

${{1 - {{64} \over {113}}} \over {1 - {{49} \over {113}}}} = {{49} \over {64}}$

(ii) Cot2 θ=(cot θ)2= 49/64

 

Question 8 

If 3 cot A = 4, check whether below is true or not

${{1 - ta{n^2}A} \over {1 + ta{n^2}A}} = co{s^2}A - si{n^2}A$

Solution 8

Given

Cot A=4/3

Or

B/P=4/3

Let B=4k and P=3k

question 8 image

So in a right angle triangle with angle A

P2 + B2 =H2

H=5k

Now tan A=1/cot A=3/4

Cos A=B/H=4/5

Sin A=P/H=3/5

Let us take the LHS

${{1 - ta{n^2}A} \over {1 + ta{n^2}A}} = {{1 - {{\left( {{3 \over 4}} \right)}^2}} \over {1 + {{\left( {{3 \over 4}} \right)}^2}}} = {7 \over {25}}$

RHS=cos2 A -  sin2A= 7/25

So LHS=RHS,so the statement is true

Question 9

In triangle ABC, right-angled at B, if $\tan A = {1 \over {\sqrt 3 }}$

Find the value of:

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

(ii) cos A cos C – sin A sin C

Solution

$\tan A = {1 \over {\sqrt 3 }}$

${P \over B} = {1 \over {\sqrt 3 }}$

Let $P = k$ and $B = k\sqrt 3 $

Now by Pythagoras theorem

P2 + B2 =H2

H=2k

(i)

$SinACosC + CosASinC = \left( {{P \over H}} \right)\left( {{B \over H}} \right) + \left( {{B \over H}} \right)\left( {{P \over H}} \right)$

$ = \left( {{{BC} \over {AC}}} \right)\left( {{{BC} \over {AC}}} \right) + \left( {{{AB} \over {AC}}} \right)\left( {{{AB} \over {AC}}} \right)$

$ = {{{k^2}} \over {4{k^2}}} + {{3{k^2}} \over {4{k^2}}} = 1$

(ii)

$CosACosC - SinASinC = \left( {{P \over H}} \right)\left( {{P \over H}} \right) + \left( {{B \over H}} \right)\left( {{B \over H}} \right)$

$ = \left( {{{BC} \over {AC}}} \right)\left( {{{AB} \over {AC}}} \right) - \left( {{{AB} \over {AC}}} \right)\left( {{{BC} \over {AC}}} \right)$

$=0$

Question 10

In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of

sin P, cos P and tan P.

Solution

Let QR=x andPR=y

Then x+y=25

y=25-x

Now by Pythagorus theorem

x2  + 25= y2

x2  + 25==(25-x)2

Solving it ,we get

X=12 cm

Then y=25-12=13 cm

Now Sin P= 12/13

Cos P=5/13

Tan P=12/5

 

Question 11

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.

(iv) cot A is the product of cot and A.

(v) sin θ =4/3 for some angle

Solution 

  1. False. The value of  tan A increase from 0 to ∞.
  2. True. The value of sec A increase from 1 to ∞.
  3. False .Cos A is the abbreviation used for the cosine of angle A
  4. False .cot A is one symbol. We cannot separate it
  5. False. The value of sin θ always lies between 0 and 0 and 4/3 > 1
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