Question

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Answer 1

Question :-

In a triangle ABC if angle B = 5 angle A and angle A = 3 angle C, find the angles of the triangles.

Given, $\angle B = 5 \angle C$

$\implies \angle C = \frac{\angle B}{5}$

And, $\angle A = 3 \angle C$

$\implies angle A = 3 \frac{\angle B}{5}$

since, angle C = angle B/5

Now, by angle sum property of triangle,

A + B + C = 180°

→

$\frac{3b}{5} + b \: + \frac{b}{5} = 180 \\ \\ \implies \: \frac{4b}{5} + \frac{5b}{5} = 180 \\ \\ \implies \: \frac{9b}{5} = 180 \\ \\ \implies \: b \: = 180 \times \frac{5}{9} \\ \\ \implies \: b \: = 100 {}^{ \circ}$

Now, angle B = 100°

So, angle A = 3/5 × 100

→ Angle A = 60°

angle C = 100/5

→ angle C = 20°