Question

In the given figure, AOB = 90 and ABC = 30 then find CAO = ?

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Answer 1

In ∆AOB

AOB + AOB + OBA = 180°

OAB = OBA

So,

2OAB + 90° = 180°

2OAB = 180° - 90°

OAB = 90/2

OAB = 45°

The angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.

So,

$\mathtt{ACB= \frac{1}{2} \times AOB } \\ \\ \mathtt{ = \frac{90}{2} } \\ \\ \implies45\degree$

In ∆ABC

ACB + CBA + BAC = 180°

45° + 30° + BAC = 180°

BAC = 180° - 75°

BAC = 105°

Now,

CAO = BAC − OAB

CAO = 105° - 45°

CAO = 60° ans