Question

In the given diagram an isosceles ΔABC is inscribed in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and ∠COM = 65°.

Find:

1. ∠ABC
2. ∠BAC
3. ∠BCQ

Answer 1

PQ is tangent to circle OM is perpendicular PQ chord AC, and ∠COM = 65°

a. Here, ∠AOM = ∠COM = 65°

= 65° + 65°

= 130°

Now, ∠ABC = `1/2` ∠AOC  ...(Since, the angle at the centre is twice the angle formed by the same arc at any other point of the circle)

= `1/2 xx 130^circ`

= 65°

b. In ΔABC,

AB = AC

∠ABC = ∠ACB = 65°  ...(Since, angles opposite to equal sides are equal)

∴ ∠BAC = 180° – (65° + 65°)

= 180° – 130°

= 50°

c. ∠OCQ = 90°  ...(Since, angle between the radius and the tangent is 90°) 

In ΔOMC,

∠OCM = 180° – (∠OMC + ∠MOC)   ...[By the angle sum property of a triangle]

= 180° – (90° + 65°)

= 180° – 155°

= 25°

∠ACB = 65°

∠OCB = ∠ACB – ∠OCM

= 65° – 25°

= 40°

∠BCQ = ∠OCQ – ∠OCB

= 90° – 40°

= 50°