Answers

2016-03-21T14:45:03+05:30
Given
∠PCA = 110°
PC is the tangent to the circle whose centre is O.
Construction
Join points C and O.

∠BCA = 90° [Since angle in a semi circle is 90°]
Also ∠PCO = 90° [Since radius ⊥ tangent]

From the figure we have,
∠PCA =∠PCO + ∠OCA 
i.e. 110° = 90° + ∠OCA 
Therefore, ∠OCA =20°

Now in ΔAOC, AO = OC [Radii] 
So, ∠OCA = ∠OAC =20°

In ΔABC, we have
∠BCA = 90° & ∠CAB = 20° 
Therefore,  ∠CBA = 70°
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