Circle

Exercise 10.5 Part 1

Question 1: In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Answer: ∠AOC = 60° + 30° = 90°
Since angle made by an arc on the centre is double the angle made anywhere else on the circle
Hence, ∠ADC = 90°/2 = 45°

Question 2: A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer: Given a circle with centre O in which chord AB = radius OB

As OA = OB = AB
So, ΔOAB is equilateral triangle
So, ∠AOB = 60°
Angle made on centre by an arc is double the angle made anywhere else on the circle
Hence, ∠ADB = 30°
As ACBD is a cyclic quadrilateral
Hence, ∠ACB = 180°- ∠ADB
Or, ∠ACB = 180°- 30° = 150°

Question 3: In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Answer: Here, PQRS is a cyclic quadrilateral

Hence, ∠PSR = 180° - ∠PQR
Or, ∠PSR = 180° - 100° = 80°
As angle made by an arc on centre is double than angle made anywhere else
So, ∠POR = 2 x 80° = 160°
In ΔPOR we have PO = RO
So, ∠OPR = ∠ORP
Or, 160° + 2∠OPR = 180°
Or, 2∠OPR = 180° - 160°= 20°
Or, ∠OPR = 10°

4. In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Answer:In ΔABC;
∠BAC = 180° - (69° + 31°)
Or, ∠BAC = 180° - 100° = 80°
Angles made on one side of a chord are equal
Hence, ∠BDC = ∠BAC = 80°