Question

1. In Fig. 10.36, A,B and C are three points on a
circle with centre O such that Z BOC = 30° and
Z AOB = 60°. If D is a point on the circle other
than the arc ABC, find ZADC.​

Answer 1

${\huge{\underline{\small{\bold{\red{Answer:-}}}}}}$

${\huge{\underline{\small{\bold{\orange{Given:-}}}}}}$

$⇝$∠AOB = 60°

$⇝$∠BOC= 30°

${\huge{\underline{\small{\bold{\pink{Here,}}}}}}$

$⇝$∠AOC = ∠AOB + ∠BOC

$⇝$∠AOC = 60°+30°

$⇝$∠AOC = 90°

${\huge{\underline{\small{\bold{\orange{Since,}}}}}}$

Arc ABC makes an angle of 90° at the centre of the circle.

$⇝$∠ADC= 1/2∠AOC

[SINCE THE ANGLE SUBTENDED BY AN ARC AT THE CENTRE IS DOUBLE THE ANGLE SUBTENDED BY IT AT ANY POINT ON THE REMAINING PART OF THE CIRCLE]

$⇝$∠ADC= 1/2(90°)

$⇝$∠ADC= 45°

${\huge{\underline{\small{\bold{\blue{Hence,}}}}}}$

$⇝$ ∠ADC= 45°

${\huge{\underline{\sf{\bold{\red{Hope\:help\:u:)}}}}}}$

Answer 2

Given :

• ∠BOC = 30°
• ∠AOB = 60°

To Find :

• ∠ADC

Solution :

$\longmapsto\tt{\angle{AOC}=\angle{AOB}+\angle{BOC}}$

$\longmapsto\tt{\angle{AOC}=60\degree+30\degree}$

$\longmapsto\tt\bf{\angle{AOC}=90\degree}$

As we know that The angle subtended by an arc on centre is double of the angle subtended by it on the other part of the circle . So ,

$\longmapsto\tt{\angle{ADC}=\dfrac{1}{2}\times{\angle{AOC}}}$

$\longmapsto\tt{\angle{ADC}=\dfrac{1}{2}\times{{\cancel{90\degree}}}}$

$\longmapsto\tt\bf{\angle{ADC}=45\degree}$

So , The Measure of ∠ADC is 45°...