In this post, you will find the NCERT solutions for class 10 maths ex 8.1. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 8.1
1. In ∆ABC, right-angled at B, AB = 24
cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution: Consider a right-angled triangle ABC,
right-angled at B.
Using
Pythagoras theorem, we have
AC2
= AB2 + BC2
= (24)2 + (7)2 =
576 + 49 = 625
Taking
square root of both side, we get
AC
= 25 cm
(i) sin A = BC/AC =
7/25, cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25,
cos C = BC/AC = 7/25
2. In the following figure,
find tan P – cot R:
Solution: Using Pythagoras theorem, we have
PR2
= PQ2 + QR2
⇒ (13)2 = (12)2 +
QR2
⇒ QR2
=169 – 144 = 25
Taking
square root of both side, we get
QR
= 5 cm
Therefore, tan P – cot R = QR/PQ –
QR/PQ = 5/12 – 5/12 = 0
3. If sin A = ¾, calculate cos A and tan A.
Solution: Consider ABC is a triangle right-angled at B.
It is given that sin A = ¾. In ∆ABC,
sin A = BC/AC.
Let
BC = 3k and AC = 4k
Then,
Using Pythagoras theorem, we have
AB2
+ BC2 = AC2
AB2
= AC2 – BC2
AB2
= (4k)2 – (3k)2
AB2
= 16k2 – 9k2
AB2
= 7k2
Taking
square root of both side, we get
AB = k√7
Therefore, cos A = AB/AC = k√7/4k = √7/4
And
tan A = BC/AB = 3k/ k√7 = 3/√7
4. Given 15 cot A = 8, find sin A and sec A.
Solution: Consider ABC is a triangle right-angled
at B.
It is given that 15 cot A = 8
⇒ cot A = 8/15
And
cot A = AB/BC
Let
AB = 8k and BC = 15k
Then
using Pythagoras theorem, we have
AC2
= AB2 + BC2
= (8k)2 + (15k)2
= 64k2 + 225k2
AC2 = 289k2
Taking
square root of both side, we get
AC
= 17k
Therefore, sin A = BC/AC = 15k/17k = 15/17
And sec A = AC/AB = 17k/8k = 17/8
5. Given sec θ
= 13/12, calculate all other trigonometric
ratios.
Solution: Consider a triangle ABC in which ∠A = θ
and ∠B = 90°.
It
is given that sec θ = 13/12 and sec θ = AC/AB.
Let
AC = 13k and AB = 12k
Then,
using Pythagoras theorem, we have
BC2
= AC2 – AB2
= (13k)2 – (12k)2
= 169k2 – 144k2
= 25k2
BC2
= 25k2
Taking
square root of both side, we get
BC
= 5k
Therefore, sin θ = BC/AC = 5k/13k = 5/13
cos θ = AB/AC = 12k/13k = 12/13
tan θ = BC/AB = 5k/12k = 5/12
cot θ = AB/BC = 12k/5k = 12/5
cosec θ = AC/BC = 13k/5k = 13/5
6. If ∠A and ∠B
are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution: In right-angled ∆ABC, ∠A and ∠B are acute angles.
Then
cos A = AC/AB and cos B = BC/AB
But cos A = cos B [Given]
⇒
AC/AB = BC/AB
⇒ AC = BC
⇒ ∠A
= ∠B [Angles opposite to
equal sides are equal.]
7. If cot θ
= 7/8, evaluate:
(i) 
(ii) cot2 θ
Solution: Consider a triangle ABC in which ∠A = θ and ∠B = 90°.
It is given that cot θ = 7/8 and from the figure we have
cot θ = AB/BC.
Let
AB = 7k and BC = 8k
Then,
using Pythagoras theorem, we have
AC2
= AB2 + BC2
AC2
= (7k)2 + (8k)2
= 49k2 + 64k2
AC2
= 113k2
Taking
square root of both side, we get
AC
= k√113
Therefore, sin θ = BC/AC = 8k/k√113 = 8/√113
Cos
θ = AB/AC = 7k/k√113
(i) 
= 
or
not.Solution: Consider a triangle ABC in which ∠B = 90°.
It
is given that 3 cot A = 4
⇒ cot A = 4/3
Let
AB = 4k and BC = 3k
Then,
using Pythagoras theorem, we have
AC2
= AB2 + BC2
AC2
= (4k)2 + (3k)2
AC2
= 16k2 + 9k2
AC2
= 25k2
Taking
square root of both sides, we get
AC
= 5k
Therefore, sin A = BC/AC = 3k/5k = 3/5
Cos A = AB/AC = 4k/5k = 4/5
And
tan A = BC/AB = 3k/4k = 3/4
Now,
L.H.S. = 
R.H.S.
= cos2 A – sin2 A =
(4/5)2 – (3/5)2
=
16/25 – 9/25 = 7/25
Since, L.H.S.
= R.H.S.
Therefore,
9. In ∆ABC 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
Solution: Consider a triangle ABC in which ∠B = 90°.
It
is given that tan A = 1/√3 and from the figure tan A = BC/AB
Let
BC = k and AB = √3k
Then,
using Pythagoras theorem, we have
AC2
= AB2 + BC2
AC2
= (√3k)2 + (k)2
AC2
= 3k2 + k2
AC2
= 4k2
Taking
square roots of both sides, we have
AC
= 2k
Therefore, sin A = BC/AC = k/2k = 1/2
And
cos A = AB/AC = √3k/2k = √3/2
For ∠C, base = BC, perpendicular = AB and hypotenuse
= AC
Therefore, sin C = AB/AC = √3k/2k = √3/2
And
cos A = BC/AC = k/2k = 1/2
(i) sin
A cos C + cos A sin C = ½ × ½ + √3/2 × √3/2
= ¼
+ ¾ = 4/4 = 1
(ii) cos A cos C – sin A sin C = √3/2 × ½ – ½ × √3/2
= √3/4
– √3/4 = 0
10. In ∆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: In ∆PQR, right-angled at Q, it is given that PR + QR = 25 cm and PQ =
5 cm.
Let
QR = x cm
Then,
PR = 25 – QR
PR
= (25 – x) cm
Using
Pythagoras theorem, we have
PR2
= QR2 + PQ2
⇒ (25 – x)2 = (x)2
+ (5)2
⇒ 625 – 50x + x2 = x2 + 25
⇒ –50x = –600
⇒ x = 12
Therefore, QR
= 12 cm and PR = 25 – 12 = 13 cm
So, sin P = QR/PR = 12/13
cos
P = PQ/PR = 5/13
And tan
P = QR/PQ = 12/5
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 θ.
Solution:
(i) False, because
sides of a right-angled triangle may have any length, so tan A may have any value.
(ii) True, because sec
A is always greater than 1.
(iii) False, because cos A is the
abbreviation of cosine A.
(iv)
False, because cot
A is not the product of ‘cot’ and A. ‘cot’ is separated from A has no
meaning.
(v) False, because sin θ cannot be greater
than 1.