Question

Ac is the diameter of the circle with center o and A is point of contact ,then find x​

Answer 1

Answer:

GIVEN:

∠BAQ = 40°

OA ⟂ PQ

∠CAQ = 90°

[We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.]

∠CBA = 90°  [angle in a semicircle is a right angle]

∠CAQ = ∠CAB + ∠BAQ

90° =  ∠CAB + 40°

∠CAB = 90° - 40° = 50°

In ∆ABC,

∠CAB + ∠CBA + ∠BCA = 180°

[Angle sum property]

50° + 90° + x = 180°

140° + x = 180°

x = 180° - 140°

x = 40 °

Hence, the value of x is 40°.

HOPE THIS WILL HELP YOU..