Question

In the given figure triangle ABC is circumscribed in a circle of radius 10 cm. The value of $\frac{AB}{sinC}$ is equal to

• A. 20 cm
• B. 10 cm
• C. 40 cm
• D. $20\sqrt{2}cm$

Answer 1

The correct option is A

20 cm

Extend the radius to form diameter BD and Join CD.

In triangle BDC,

$sinD=\frac{BC}{BD}=\frac{BC}{20}$

Therefore, $\frac{BC}{sinD}=20$

$\Rightarrow \frac{BC}{sinA}=20$......(i) ($\angle A=\angle D$ as they are angles subtended by the same chord BC)

Applying sine rule triangle ABC, we get

$\frac{AB}{sinC}=\frac{BC}{sinA}=\frac{AC}{sinB}$......(ii)

From (i) and (ii), we get

$\frac{AB}{sinC}=20cm$