Question

In the figure below, one angle of a triangle is ${120}^{\circ }$ and its opposite side is 4cm. The diameter of its circumcenter will be equal to

• A. $\frac{8}{\sqrt{3}}cm$
• B. $\frac{8}{\sqrt{6}}cm$
• C. $\frac{116}{\sqrt{3}}cm$
• D. $\frac{4}{\sqrt{3}}cm$

Answer 1

The correct option is A

$\frac{8}{\sqrt{3}}cm$

In $\Delta ABC$

$\angle B={120}^{\circ },AC=4cm$

Draw diameter AD and join CD to form a quadrilateral ABCD

$\angle ADC=180-{120}^{\circ }={60}^{\circ }$ (ABCD is a cyclic quadrilateral )

$\angle ACD={90}^{\circ }$ (Angle subtended by the diameter on the circumference)

Angles of $\Delta ADC$ are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

Corresponding sides can be calculated as

$\Rightarrow \sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{2}:\frac{\sqrt{3}}{2}:1$

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ x& :& x\sqrt{3}& :& 2x\\ \downarrow & & \downarrow & & \downarrow \\ CD& & AC& & AD\\ & & \downarrow & & \\ \frac{4}{\sqrt{3}}& & 4cm& & \frac{8}{\sqrt{3}}cm\end{array}$

So, the diameter $AD=\frac{8}{\sqrt{3}}cm.$