The tangent at a point C of a circle with centre O and a diameter AB, which when extended intersect at point P. If ∠PCA = 110º , find ∠CBA and ∠CPA . [ pls solve with full steps ]
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Answer: 70°
Step-by-step explanation:
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° [angle sum property]
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