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