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Rotation Concept

The sides of ...

Question

The sides of a triangle are $3 x + 4 y, 4 x + 3 y$ and $5 x + 5 y$ units, where $x > 0, y > 0.$ then prove that the triangle is obtuse-angled.

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Solution

Let $a = 3 x + 4 y; b = 4 x + 3 y$ and $c = 5 x + 5 y$
$∴ cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
$= \frac{{(3 x + 4 y)}^{2} + {(4 x + 3 y)}^{2} - {(5 x + 5 y)}^{2}}{2 (3 x + 4 y) (4 x + 3 y)}$
$= \frac{9 x^{2} + 16 y^{2} + 24 x y + 16 x^{2} + 9 y^{2} + 24 x y - 25 x^{2} - 25 y^{2} - 50 x y}{2 (3 x + 4 y) (4 x + 3 y)}$
$= \frac{- 2 x y}{2 (3 x + 4 y) (4 x + 3 y)} < 0$ ( $∵ x > 0, y > 0$ )
$\Rightarrow cos C < 0$
$∴ △$ is obtuse angled.

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