Question

If A, B, C are three points on a circle with centre O such that $\angle AOB={90}^{o}and\angle BOC={120}^o,then\angle ABC=$

• A. ${60}^o$
• B. ${75}^o$
• C. ${90}^o$
• D. ${135}^o$

Answer 1

The correct option is B

${75}^o$

A, B and C are three points on a circle with centre O.

$\angle AOB={90}^{o}and\angle BOC={120}^o$

$\therefore \angle AOC={360}^o-({120}^o+{90}^o)$$={360}^o-{210}^o={150}^o$

But $\angle AOC$ is at the centre made by are AC and $\angle ABC$ at the remaining part of the circle.

$\therefore \angle ABC=\frac{1}{2}\angle AOC$$=\frac{1}{2}\times {150}^o={75}^o$