# Thread:

1. ## Find the angles of the triangle?

Each of the two triplets of numbers $(a, b, c)\ \& \left(\frac{a}{2b},\ \frac{2b}{3c},\ \frac{3c}{a}\right)$ is a Geometric Progression. Can the numbers $a,b,c$ be the lengths of the sides of a triangle? If they can what triangle is it? Find the angles of the triangle.

2. Originally Posted by fardeen_gen
Each of the two triplets of numbers $(a, b, c)\ \& \left(\frac{a}{2b},\ \frac{2b}{3c},\ \frac{3c}{a}\right)$ is a Geometric Progression. Can the numbers $a,b,c$ be the lengths of the sides of a triangle? If they can what triangle is it? Find the angles of the triangle.
Use the fact that they are in a GP, the two equations ( $b^2 = ac$ and $\frac{4b^2}{9c^2} = \frac{a}{2b}\frac{3c}{a}$)to get $9c = 4a = 6b$. They clearly form a triangle. As for the angles, compute them for the triangle with sides 4,6,9.
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