Question

The tangent at a point C on the circle and diameter BA when extended intersect at P such that angle PAC = 10° find angle CBA

Answer 1

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°

Answer 2

Answer:

Step-by-step explanation: the simplest and step -by -step!