Question

Two chords $AB$ and $AC$ of a circle subtends angles equal to $90°$and $150°$, respectively at the centre. Find $\angle BAC$, if $AB$ and $AC$ lie on the opposite sides of the centre.

Answer 1

• Step 1: Let's find $\angle OAB$

In $∆AOB$,

$OA=OB$ (radius of the circle)

Since angle opposite to equal sides are equal, we obtain

$\angle OBA=\angle OAB$

We know that,

According toangle sum property of triangle theorem , sum of all angles of a triangle $=180°$

As per the angle sum property in $△AOB$, we get,

$\begin{array}{rcl}\angle OAB+\angle AOB+\angle OBA& =& 180°\\ & \Rightarrow & \angle OAB+90°+\angle OAB=180°\\ & \Rightarrow & 2\angle OAB=90°\\ & \Rightarrow & \angle OAB=\frac{90°}{2}\\ & \Rightarrow & \angle OAB=45°\end{array}$

• Step 2: Let's find $\angle OAC$

Now, in $∆AOC,$

$OA=OC$ (radius of the circle)

Since, angle opposite to equal sides are equal

∴ $\angle OCA=\angle OAC$

Using the angle sum property in $∆AOB$, sum of all angles of the triangle is $180°$, we have:

$\begin{array}{rcl}\angle OAC+\angle AOC+\angle OCA& =& 180°\\ & \Rightarrow & \angle OAC+150°+\angle OCA=180°\\ & \Rightarrow & 2\angle OAC=180°-150°\\ & \Rightarrow & 2\angle OAC=30°\\ & \Rightarrow & \angle OAC=15°\end{array}$

Now,

$\begin{array}{rcl}\angle BAC& =& \angle OAB+\angle OAC\\ & =& 45°+15°\\ & =& 60°\\ \angle BAC& =& 60°\end{array}$

Hence, $\angle BAC=60°$