Question

In $\Delta ABC,\angle A={40}^{\circ },BC=8cm.$ Find the circumdiameter of the triangle. $[sin{40}^{\circ }=0.64]$

• A. 10.5 cm
• B. 12.5 cm
• C. 14 cm
• D. 15 cm

Answer 1

The correct option is B

12.5 cm

Draw diameter BD passing through centre O, join CD

In $\Delta BCD,\angle D={40}^{\circ }$ (angle subtended by the same chord BC)

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

In right - angled triangle BCD

$sin{40}^{\circ }=\frac{BC}{BD}=\frac{8}{BD}$

$BD=\frac{8}{sin{40}^{\circ }}=\frac{8}{0.64}=12.5cm$

Hence, circumdiameter = 12.5 cm