Question

PC is tangent to circle with centre O at C. AOB is the diameter when extended meets tangent at P. Find angle CBA given that angle PCA = 130o. Then also find angle AOC and angle BCO

Answer 1

Given :- PC is tangent to circle with centre O at C. AOB is the diameter when extended meets tangent at P. Find angle CBA given that angle PCA = 130°. Then also find angle AOC and angle BCO ?

Solution :-

from image we have,

→ ∠PCA = 130° (given.)

→ ∠ACB = 90° (Angle in semi circle.)

→ ∠OCP = 90° (given PC is a tangent , so, by tangent property.)

Than,

→ ∠PCA = ∠OCP + ∠ACO

→ 130° = 90° + ∠ACO

→ ∠ACO = 130° - 90° = 40°

Therefore,

→ ∠BCO = ∠ACB - ∠ACO = 90° - 40° = 50° (Ans.)

Now,

→ ∠PCA = ∠ACB + ∠BCP

→ 130° = 90° + ∠BCP

→ ∠BCP = 40°

So,

→ ∠BCP = ∠CAB = 40° (Angle between tangent and a chord is equal to the inscribed angle on opposite side of the chord.)

Now, in ∆AOC, we have,

→ ∠ACO + ∠CAO + ∠ AOC = 180° (Angle sum property).

→ 40° + 40° + ∠AOC = 180°

→ ∠AOC = 180° - 80° = 100° (Ans.)

Similarly,

In ∆ABC, we have,

→ ∠CAB + ∠ACB + ∠CBA = 180°

→ 40° + 90 + °∠CBA = 180°

→ ∠CBA = 180° - 130° = 50° (Ans.)

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