Question

One angle of a $\Delta ABC$ is ${150}^{\circ }$ and its opposite side is 3 cm as shown in the figure. The diameter of its circumcircle is equal to

• A. $6\sqrt{3}cm$
• B. $6\sqrt{2}cm$
• C. 6 cm
• D. $3\sqrt{3}cm$

Answer 1

The correct option is C

6 cm

In $\Delta ABC$

$\angle B={150}^{\circ },AC=3cm$

Draw diameter AD and join CD to form quadrilateral ABCD

$\angle ADC={180}^{\circ }-150={30}^{\circ }$ (ABCD is the cyclic quadrilateral)

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

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

The corresponding sides can be calculated as

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

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

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

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ x& :& x\sqrt{3}& :& 2x\\ \downarrow & & \downarrow & & \downarrow \\ AC& & CD& & AD\\ \downarrow & & \downarrow & & \downarrow \\ 3cm& & 3\sqrt{3}cm& & 6cm\end{array}$

Hence diameter AD = 6 cm