Question

In $\Delta ABC,\angle A={50}^{\circ },BC=8cm.$ Find the circumdiameter of the triangle $[sin{50}^{\circ }=0.77]$

• A. 9.8 cm
• B. 12.54 cm
• C. 10.38 cm
• D. 11.4 cm

Answer 1

The correct option is C

10.38 cm

Draw diameter BD and connect DC

In $\Delta BCD,\angle D={50}^{\circ },$ (Angle subtended by the same chord BC)

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

In $\Delta BCD$

$sin{50}^{\circ }=\frac{BC}{BD}=\frac{8}{BD}$

$BD=\frac{8}{sin{50}^{\circ }}=\frac{8}{0.77}=10.38cm$