A circle is drawn on AB as diameter the centre of the circle is o and the length of AB is is 13 cm. P is a point on the circumference of the circle such that the chord AP equals to 12 CM. 1. Value of the PAB and POB in radian. 2. The area of the sector bounded by OP,OB and the miner arc PB.
Answers
Given:
A circle is drawn on AB as diameter the centre of the circle is o and the length of AB is is 13 cm. P is a point on the circumference of the circle such that the chord AP equals to 12 CM.
To find:
1. Value of the PAB and POB in radian.
2. The area of the sector bounded by OP,OB and the miner arc PB.
Solution:
From given, we have,
A circle is drawn on AB as diameter the centre of the circle is o and P is a point on the circumference of the circle.
⇒ ∠ APB = 90° (The angle APB is an angle in a semicircle.)
In Δ APB, using Pythagoras theorem, we get
AB² = AP² + PB²
⇒ 13² = 12² + PB²
⇒ 169 = 144 + PB²
∴ PB = 5 cm
1. Value of the PAB and POB in radian.
Now consider,
tan ∠ PAB = PB/PA
∠ PAB = tan^{-1} (PB/PA)
∠ PAB = tan^{-1} (5/12)
∴ ∠ PAB = 0.3947 rad
AS ∠ PAB is an angle at the circumference subtended by the chord PB, so we have,
∠ POB = 2 × ∠ PAB
∠ POB = 2 × 0.3947
∴ ∠ POB = 0.7894 rad
2. The area of the sector bounded by OP,OB and the miner arc PB.
Area of sector POB = 1/2 × OB² × ∠ POB
= 1/2 × (13/2)² × 0.7894
= 16.67 cm²
∴ Area of sector POB = 16.67 cm²