Question
In the given figure, AB is the diameter of a circle with
a center O, and AT is a tangent. If ∠ AOQ=58 ∘ , find ∠ ATQ.Solution
AB is the straight line. ∠AOQ +∠BOQ =180 ° ∠BOQ =180 ° -58 ° =122 ° In triangle BOQ, OB and OQ are equal. Since they are radius (OB=OQ) So, ∠OBQ=∠OQB ∠OBQ+∠OQB+∠BOQ =180 ° 122+2(∠OBQ) =180 ° ∠OBQ=29 ° In the triangle ABT – ∠ABT+∠BAT+∠BTA =180 ° =29 ° +90 ° +∠BAT =180 ° ∠BAT =180-119=61 ° ∠ ATQ=61°
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