Question

please answer this question 8...

Answer 1

So, Here is ur answer:- Given ∠PCA = 110° and PC is the tangent. Given O is the centre of the circle. Hence points A, O, B and P all lie on the same line. Join points C and O. ∠BCA = 90° [Since angle in a semi circle is 90°] Also ∠OCP = 90° [Since radius ⊥ tangent] From the figure we have, ∠PCA =∠PCO + ∠OCA That is, 110° = 90° + ∠OCA Therefore, ∠OCA =20° In ΔAOC, AO = OC [Radii] So, ∠OCA = ∠OAC =20° In ΔABC, we have ∠BCA = 90°, ∠CAB = 20° Therefore, ∠CBA = 70° Hope it helps!!

Answer 2

Answer:

Step-by-step explanation:

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] [1]

Also,

∠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° [2]

Now In △ABC By angle sum property of Triangle.

∠CBA + ∠BAC + ∠ACB = 180

∠CBA + 20 + 90 = 180

∠CBA = 70°