Question

AB is the diameter of circle with centre O and AC is a tangent. If angle AOD=60 degree, find angle ACD

Answer 1

AB is the straight line 

∠AOQ+∠BOQ=180°

∠BOQ=180°−58

∠BOQ=122°

In triangle BOQ,OB+OQ are equal since they are radius of the circle (OB=OQ)

So ∠OBQ=∠OQB (Since sides opposite are equal angle opposite to the equal sides are equal)

So ∠OBQ+∠OQB+∠BOQ=180°

122°+2(∠OBQ)=180°→∠OBW=29°

In triangle ABT⇒∠ABT+∠BAT+∠BTA=180o=29°+90°+∠BAT=180°

∠ATQ=61°

Step-by-step explanation:

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