Question

Prove that the angle in a segment greater than a semi-circle is less than a right angle.

Answer 1

Given: A segment ACB is greater than a semicircle.

To prove: $\angle ACB<{90}^o$

Construction: Join OA and OB.

Proof: Arc ADB subtends $\angle AOB$ at the centre and $\angle ACB$ at the remaining part of the circle.

$\therefore \angle ACB=\frac{1}{2}\angle AOB$

But $\angle AOB<{180}^o$ (A straight angle)

$\therefore \angle ACB<\frac{1}{2}\times {180}^o$

$\Rightarrow \angle ACB<{90}^o$

Hence, $\angle ACB<{90}^o$