Question

1. In Fig. 15 . 120 . O is the centre of the circle of angle APB = 50° . Find angle AOB and angle OAB . ​

Answer 1

➩Given

→A circle with centre O

→∠APB=50°

➩To Find

(i)∠AOB

(ii)∠OAB

➩Solution

(i)∠AOB

∠AOB=2∠APB (Angle subtended by an arc is double the angle subtended by the same arc at the remaining part of a circle○ )

∠APB=50°

∠AOB=2∠APB

∠AOB=2×50°

∠AOB=100°

(ii)∠OAB

We know AO=BO ( Radius of a circle are always equal)

△OAB is an isoceles triangle △ because two sides AO and BO are equal

then,

∠OAB=∠OBA(Angles opposite to equal sides are always equal)

∠OAB+∠OBA+∠AOB=180(Sum of all three angles of a triangle is 180°)

∠OAB+∠OBA+100°=180

∠OAB+∠OAB+100°=180(as ∠OAB=∠OBA)

2∠OAB+100°=180°

2∠OAB=180°-100°

2∠OAB=80°

$∠OAB=\frac{80}{2}$

∠OAB=40°

Answer 2

Solution :-

(i) ∠AOB

∠AOB=2∠APB (Angle subtended by an arc is double the angle subtended by the same arc at the remaining part of a circle○ )

➞ ∠APB = 50°

➞ ∠AOB = 2∠APB

➞ ∠AOB = 2 × 50°

∠AOB = 100°

(ii) ∠OAB

We know AO = BO ( Radius of a circle are always equal)

Then,

➞ ∠OAB = ∠OBA

➞ ∠OAB +∠OBA+∠AOB = 180

➞ ∠OAB +∠OBA + 100° = 180

➞ ∠OAB + ∠OAB + 100° = 180

➞ 2∠OAB + 100° = 180°

➞ 2∠OAB = 180° - 100°

➞ 2∠OAB = 80°

➞ ∠OAB = 80/2

∠OAB=40°