Q5 of 47 Page 1

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

∠ AOQ = 58° (given)

∠ OBQ = 1/2 ∠AOQ (angle in a semicircle)


⇒ ∠ OBQ = 1/2 × 58° = 29°


In Δ BAT,
∠ABQ + ∠BAT + ∠ATB = 180° (Sum of all the angles of a triangle)


⇒ 29° + 90° + ∠ ATB = 180°


⇒ ∠ ATB = 61°


∴ ∠ ATQ = 61°


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