Question

plzz answer this question ✋​

Answer 1

Step-by-step explanation:

B is the straight line 

∠AOQ+∠BOQ=180o

∠BOQ=180o−58

∠BOQ=122o

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=180o

122o+2(∠OBQ)=180o→∠OBW=29o

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

∠ATQ=61o