Question

In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If $\angle APB={70}^o$, find $\angle ACB$.

Answer 1

Arc AB subtends $\angle AOB$ at the centre and $\angle APB$ at the remaining part of the circle.

$\therefore \angle AOB=2\angle APB=2\times {70}^o={140}^o$

Now in cyclic quadrilateral AOBC, $\angle AOB+\angle ACB={180}^o$ (Sum of opposite angles)

$\Rightarrow {140}^o+\angle ACB={180}^o$

$\Rightarrow \angle ACB={180}^o-{140}^o={40}^o$

$\therefore \angle ACB={40}^o$