Two tangents PA and PB are drawn to a circle with centre O from an external point P. If ∠OAB is 30°.
Find angle APB.
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Answer:
the answer is 60 degree
Step-by-step explanation:
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AB is the straight line
∠AOQ+∠BOQ=180
o
∠BOQ=180
o
−58
∠BOQ=122
o
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
o
122
o
+2(∠OBQ)=180
o
→∠OBW=29
o
In triangle ABT⇒∠ABT+∠BAT+∠BTA=180
o
=29
o
+90
o
+∠BAT=180
o
∠ATQ=61
o
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