Question

$\triangle ABC$ is inscribed in a circle of radius 10 cm. If $a=10$ cm then $\angle B + \angle C = ?$

• A. $45^{\circ}$
• B. $30^{\circ}$
• C. $150^{\circ}$
• D. $120^{\circ}$

Answer 1

Answer: (C)

$\angle B + \angle C = 150^{\circ}$

Answer 2

Given: Triangle $ABC$ is inscribed in a circle of radius $R = 10$ cm and side $a = 10$ cm (opposite to angle $A$).

Using the chord formula:

$$a = 2R \sin A \implies 10 = 2 \times 10 \times \sin A \implies \sin A = \frac{10}{20} = 0.5$$

Thus, $A = 30^\circ$.

Since the sum of angles in a triangle is $180^\circ$:

$$\angle B + \angle C = 180^\circ - A = 180^\circ - 30^\circ = 150^\circ.$$