Question

In the figure, O is the centre of circle and AO is
perpendicular to BO. Find Angle CAO
​

Answer 1

Solution :-

in ∆AOB , we have ,

→ ∠AOB = 90° .(given)

→ AO = OB (radius of circle.)

So,

→ ∠OAB = ∠OBA . (Angle opposite to equal sides are equal.)

Therefore,

→ ∠AOB + ∠OAB + ∠OBA = 180° (Angle sum Property.)

→ ∠AOB + ∠OAB + ∠OAB = 180°

→ 90° + 2∠OAB = 180°

→ 2∠OAB = 180° - 90°

→ 2∠OAB = 90°

→ ∠OAB = 45° .

Now,

→ ∠ACB = (1/2)∠AOB . (Angle on the circumference is half of angle at the centre.)

→ ∠ACB = (1/2) * 90°

→ ∠ACB = 45°.

Now, in ∆ACB ,

→ ∠ACB = 45° .( solved above.)

→ ∠CBA = 30° (Given.)

→ ∠OAB = 45° .(Solved above.)

So,

→ ∠ACB + ∠CAB + ∠CBA = 180° (Angle sum Property.)

→ 45° + ∠CAB + 30° = 180°

→ ∠CAB = 180° - (45° + 30°)

→ ∠CAB = 180° - 75°

→ ∠CAB = 105°

→ ∠CAO + ∠OAB = 105°

→ ∠CAO + 45° = 105°

→ ∠CAO = 105° - 45°

→ ∠CAO = 60° .(Ans.)

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