Question

In Fig. 2, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.
 

Answer 1

We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOQ = 2∠ABQ

⇒ ∠ABQ =`1/2`∠AOQ

⇒ ∠ABQ =`1/2`×58°=29° 

or ∠ABT = 29°

We know that the radius is perpendicular to the tangent at the point of contact.

∴ ∠OAT = 90°     (OA ⊥ AT)

or ∠BAT = 90°

Now, in ∆BAT,

∠BAT+∠ABT+∠ATB=180°

⇒90°+29°+∠ATB=180°

⇒∠ATB=180°−119°=61°

∴∠ATQ=61°