Question

Given figure shows a triangle ABC circumscribing a circle. If $\angle BOC={110}^{\circ }$, then $\angle BAC=$

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${35}^{\circ }$

Answer 1

The correct option is A ${40}^{\circ }$

Let us mark the points P, Q and R where triangle is touching the circle at sides BC, AC and AB respectively.
Join the points as shown in figure.

$\because OP=OQ=OR=$ radius of circle,
$CP=CQ$ [Tangents from external point are equal]
and BP = BR [Tangents from external point are equal]

$\therefore CQOP$ and $BPOR$ are kites
Let us assume $\angle COP=\text{x}$ and $\angle BOP=\text{y}$

In kite $CQOP$ diagonal CO bisects $\angle POQ$ [property of diagonal of kite]
$\therefore \angle COQ=\angle COP=\text{x}$

Similarly, in kite BPOR,
$\angle BOR=\angle BOP=\text{y}$

Given: $\angle BOC={110}^{\circ }$
$\therefore \angle BOC=\angle BOP+\angle COP={110}^{\circ }$
$\Rightarrow \text{x}+\text{y}={110}^{\circ }$

We know that centre of circle makes an angle of ${360}^{\circ }$
$\therefore \angle BOR+\angle BOP+\angle COP+\angle COQ+\angle ROQ={360}^{\circ }$
$\Rightarrow \text{y}+\text{y}+\text{x}+\text{x}+\angle ROQ={360}^{\circ }$
$\Rightarrow 2(\text{y}+\text{x})+\angle ROQ={360}^{\circ }$
$\Rightarrow 2\left({110}^{\circ }\right)+\angle ROQ={360}^{\circ }$
$\Rightarrow \angle ROQ={360}^{\circ }-{220}^{\circ }$
$\Rightarrow \angle ROQ={140}^{\circ }$

By Theorem- Tangents at any point to the circle is perpendicular to the radius through the point of contact
$\therefore \angle ORA=\angle OQA={90}^{\circ }$
Now consider quadrilateral ORAQ,
$\therefore \angle ORA+\angle RAQ+\angle OQA+\angle ROQ={360}^{\circ }$
$\Rightarrow {90}^{\circ }+\angle RAQ+{90}^{\circ }+{140}^{\circ }={360}^{\circ }$
$\Rightarrow \angle RAQ={360}^{\circ }-{90}^{\circ }-{90}^{\circ }-{140}^{\circ }$
$\therefore \angle RAQ\left(or\right)\angle BAC={40}^{\circ }$