Question

In a triangle abc if 3∆a=4∆b=6∆c. Calculate the angle

Answer 1

Let,

3∠a=4 ∠b=6∠c = k

$= > ∠a = \frac{k}{3} \\ \\ = > ∠b = \frac{k}{4} \\ \\ = > ∠c = \frac{k}{6}$

By angle sum property; we have

∠a + ∠b + ∠c = 180°

$= > \frac{k}{3} + \frac{k}{4} + \frac{k}{6} = 180 \\ \\ = > \frac{4k + 3k + 2k}{12} = 180 \\ \\ = > {4k + 3k + 2k} = 180 \times 12 \\ \\ = > 9k = 180 \times 12 \\ \\ = > k = \frac{180 \times 12}{9} \\ \\ = > k = 240$

Hence the angles are.

$= > ∠a = \frac{k}{3} = \frac{240}{3} = 80\\ \\ = > ∠b = \frac{k}{4} = \frac{240}{4} = 60\\ \\ = > ∠c = \frac{k}{6} = \frac{240}{6} = 40$