# 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° ,

Sin A = BC/AC = 7/25
CosA = AB/AC = 24/25
(ii)

Sin C = AB/AC = 24/25
Cos C = BC/AC =7/25
Question 2
In below find tan P – cot R

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

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

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

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

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

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

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