Question

In the given figure, a circle is circumscribing $\Delta ABC$ where $\angle A={125}^{\circ }$ and side BC=8cm. The diameter of the circumcircle is equal to

$[sin{55}^{\circ }=0.82]$

• A. 12.5 cm
• B. 10.75 cm
• C. 9.75 cm
• D. 8.25 cm

Answer 1

The correct option is C

9.75 cm

Draw diameter BD of the given circle and join CD as shown in the figure.

ABCD becomes a cyclic quadrilateral.

Hence, $\angle CDB={180}^{\circ }-\left({125}^{\circ }\right)={55}^{\circ }$

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

In right-angled triangle BCD

$sin{55}^{\circ }=\frac{BC}{BD}$

$BD=\frac{BC}{sin{55}^{\circ }}=\frac{8}{0.82}=9.75cm$

So, diameter $BD=9.75cm$