Maths Class 10 Exercise 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.