Question

In the above figure, BC is a tangent at B. A is the centre of the circle. If AB = BC, then find the value of $\angle BAC$

• A. ${10}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

The correct option is C

${45}^{\circ }$

[By Theorem- The tangent at any point of a circle and the radius through this point are perpendicular to each other]
$\because$ BC is a tangent at B
$\therefore$ $\angle ABC={90}^{\circ }$

Consider $\Delta ABC,\angle B={90}^{\circ }$
Also, AB = BC [Given]

$\therefore$ $\angle BAC=\angle BCA$ [By Theorem- Angles opposite to equal sides are equal]

Let, $\angle BAC=BCA=x$

We know that, sum of angles in a triangle = ${180}^{\circ }$

$\therefore \angle ABC+\angle BCA+\angle BAC={180}^{\circ }{90}^{\circ }+x+x+{180}^{\circ }2x={180}^{\circ }-{90}^{\circ }2x={90}^{\circ }x=\frac{{90}^{\circ }}{2}x={45}^{\circ }\therefore \angle BAC=x={45}^{\circ }$