Find the angles of the triangle?

**fardeen_gen** · May 21st 2009, 10:25 AM

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.

**Isomorphism** · May 21st 2009, 10:41 AM

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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