Question

Find the value of $x$ in the following figure.

• A. ${20}^{\circ }$
• B. ${15}^{\circ }$
• C. ${10}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is B ${15}^{\circ }$

$△ABO$ is a isosceles triangle. Therefore, $\angle BAO=\angle AOB$, as the angles opposite to the equal sides are equal.

Also, $\angle AOB=\angle COD$ as they are vertically opposite angles.

Again, $△COD$ is also an isosceles triangle and hence, $\angle COD=\angle ODC$ as they are opposite to the equal sides of the triangle.

Thus, $\angle BAO=\angle AOB=\angle COD=\angle ODC$.

As the sum of three angles of a triangle is ${180}^{\circ }$,
$\angle OCD+\angle COD+\angle ODC={180}^{\circ }\Rightarrow (4x+{10}^{\circ })+(3x+{10}^{\circ })+(3x+{10}^{\circ })={180}^{\circ }\Rightarrow 10x+{30}^{\circ }={180}^{\circ }\Rightarrow 10x={180}^{\circ }-{30}^{\circ }\Rightarrow 10x={150}^{\circ }\Rightarrow x=\frac{{150}^{\circ }}{10}\Rightarrow x={15}^{\circ }$

Therefore, the value of $x$ is ${15}^{\circ }$.