Question

The perpendiculars drawn from centre O to the chords AB and AC are of equal length. Find the angle $\angle ACO.$

• A. ${20}^o$
• B. ${50}^o$
• C. ${70}^o$

Answer 1

The correct option is A ${20}^o$

Here, OB and OC are radii of the given circle.

$\therefore$ The opposite angles must be equal, i.e., $\angle OBC=\angle OCB={50}^o$

For $△BCO,$
$\angle BOC={180}^o-2\times {50}^o={80}^o$

We know that the angle subtended by an arc at the centre of a circle is double the angle subtended by the arc at any point on rest part of circumference of the circle.

Considering the minor arc BC, $\angle BAC=\frac{\angle BOC}{2}=\frac{{80}^o}{2}={40}^o$

The perpendiculars drawn from the center O to the chords AB and AC are of equal length.
$\therefore$ Chords AB and AC are of same length.

For $△ABC,$ the angles opposite to the equal sides are of same measure.
$\Rightarrow \angle ABC=\angle ACB$

From sum of interior angle property,
$\angle ABC+\angle ACB+\angle BAC={180}^o$
$\Rightarrow 2\times \angle ACB={180}^o-{40}^o={140}^o$
$\Rightarrow \angle ACB=\frac{{140}^o}{2}={70}^o$

Also, $\angle ACB=\angle OCB+\angle ACO$
$\Rightarrow \angle ACO={70}^o-{50}^o={20}^o$