Question

plz answer fast ITS VERY VERY URGENT

the tangent at a point C of a circle and diameter AB when extended intersects at P if angle PCA =110° find angle CBA​

Answer 1

Answer:

∠CBA = 70°

Step-by-step explanation:

From figure:

∠BCA = 90°

Given, ∠PCA = 110°.

⇒ ∠PCA = ∠PCB + ∠BCA

⇒ 110° = ∠PCB + 90°

⇒ ∠PCB = 20°

∴ ∠BAC = ∠PCB = 20°.

Now,

In ΔABC,

⇒ ∠CBA + ∠BCA + ∠CAB = 180°

⇒ ∠CBA + 90° + 20° = 180°

⇒ ∠CBA + 110° = 180°

⇒ ∠CBA = 70°.

Hope it helps!

Answer 2

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°