Q. 124.7( 18 Votes )
The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA.
Given : A circle with center O in which PC is a tangent a point C and AB is a diameter which is extended to P and ∠PCA = 110°
To Find : ∠CBA
∠ACB = 90° [Angle in a semicircle is a right angle] 
∠PCA = ∠ACB + ∠PCB
110 = 90 + ∠PCB
∠PCB = 20°
Now, ∠PCB = ∠BAC [angle between tangent and the chord equals angle made by the chord in alternate segment]
∠BAC = 20° 
Now In △ABC By angle sum property of Triangle.
∠CBA + ∠BAC + ∠ACB = 180
∠CBA + 20 + 90 = 180
∠CBA = 70°
Rate this question :