Find the lengths of the sides of a triangle, if its angles are in the ratio 1: 2: 3 and the circum-radius is 10 cm.

Concept:

• Sine Rule:

In a triangle Δ ABC, where a is the side opposite to A, b is the side opposite to B, c is the side opposite to C and where R is the circum-radius:

\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

Calculation:

Given: Angles of a triangle are in the ratio 1: 2: 3 and the circum-radius is 10 cm.

Let A: B: C = 1: 2: 3 or A = k, B = 2k and C = 3k where k is any real number and R = 10 cm.

 As we know that, A + B + C = 180°

⇒ A + B + C = k + 2k + 3k = 180°

⇒ k = 30°

⇒ A = 30°, B = 60° and C = 90°

As we know that, \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

\( \Rightarrow \frac{a}{{\sin 30^\circ }} = \frac{b}{{\sin 60^\circ }} = \frac{c}{{\sin 90^\circ }} = 20\)

⇒ a = 20 × sin 30° = 10 cm, b = 20 × sin 60° = 10√3 cm and c = 20 × sin 90° = 20 cm.