## 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

AC^{2}
= AB^{2} + BC^{2}

= (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

PR^{2}
= PQ^{2} + QR^{2}

⇒ (13)^{2} = (12)^{2} +
QR^{2}

⇒ QR^{2
}=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

AB^{2}
+ BC^{2} = AC^{2}

AB^{2}
= AC^{2 }– BC^{2}

AB^{2}
= (4k)^{2 }– (3k)^{2}

AB^{2}
= 16k^{2 }– 9k^{2}

AB^{2}
= 7k^{2}

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

AC^{2}
= AB^{2} + BC^{2}

= (8k)^{2} + (15k)^{2}

= 64k^{2} + 225k^{2}

AC^{2} = 289k^{2}

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 ****Î¸
= 13/12, calculate all other trigonometric
ratios.**

**Solution: **Consider a triangle ABC in which ∠A = Î¸
and ∠B = 90°.

It
is given that sec Î¸ = 13/12 and sec Î¸ = AC/AB.

Let
AC = 13k and AB = 12k

Then,
using Pythagoras theorem, we have

BC^{2}
= AC^{2} – AB^{2}

= (13k)^{2} – (12k)^{2}

= 169k^{2} – 144k^{2}

= 25k^{2}

BC^{2
}= 25k^{2}

Taking
square root of both side, we get

BC^{
}= 5k

Therefore, sin Î¸ = BC/AC = 5k/13k = 5/13

cos Î¸ = AB/AC = 12k/13k = 12/13

tan Î¸ = BC/AB = 5k/12k = 5/12

cot Î¸ = AB/BC = 12k/5k = 12/5

cosec Î¸ = 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 ****Î¸
= 7/8, evaluate:**

**(ii) **cot^{2} Î¸

**Solution: **Consider a triangle ABC in which ∠A = Î¸ and ∠B = 90°.

It is given that cot Î¸ = 7/8 and from the figure we have
cot Î¸ = AB/BC.

Let
AB = 7k and BC = 8k

Then,
using Pythagoras theorem, we have

AC^{2}
= AB^{2} + BC^{2}

AC^{2}
= (7k)^{2} + (8k)^{2}

= 49k^{2} + 64k^{2}

AC^{2}
= 113k^{2}

Taking
square root of both side, we get

AC
= k√113

Therefore, sin Î¸ = BC/AC = 8k/k√113 = 8/√113

Cos
Î¸ = AB/AC = 7k/k√113

**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

AC^{2}
= AB^{2} + BC^{2}

AC^{2}
= (4k)^{2} + (3k)^{2}

AC^{2}
= 16k^{2} + 9k^{2}

AC^{2}
= 25k^{2}

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

R.H.S.
= cos^{2} A – sin^{2} A =
(4/5)^{2} – (3/5)^{2}

=
16/25 – 9/25 = 7/25

Since, L.H.S.
= R.H.S.

**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

AC^{2}
= AB^{2} + BC^{2}

AC^{2}
= (√3k)^{2} + (k)^{2}

AC^{2}
= 3k^{2} + k^{2}

AC^{2}
= 4k^{2}

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

PR^{2}
= QR^{2} + PQ^{2}

⇒ (25 – *x*)^{2} = (*x*)^{2}
+ (5)^{2}

⇒ 625 – 50*x* + *x*^{2} = *x*^{2} + 25

⇒ –50*x* = –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 ****Î¸ = 4/3 for
some angle Î¸**.

**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 Î¸ cannot be greater
than 1.