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Question

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


Solution

Let $$a=3x+4y;    b=4x+3y$$ and $$c=5x+5y$$ 
$$\therefore \cos{C}=\dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$$
            $$=\dfrac{{\left(3x+4y\right)}^{2}+{\left(4x+3y\right)}^{2}-{\left(5x+5y\right)}^{2}}{2\left(3x+4y\right)\left(4x+3y\right)}$$
            $$=\dfrac{9{x}^{2}+16{y}^{2}+24xy+16{x}^{2}+9{y}^{2}+24xy-25{x}^{2}-25{y}^{2}-50xy}{2\left(3x+4y\right)\left(4x+3y\right)}$$
           $$=\dfrac{-2xy}{2\left(3x+4y\right)\left(4x+3y\right)}<0$$    ($$\because x>0,y>0$$)
$$\Rightarrow \cos{C}<0$$
$$\therefore \triangle$$ is obtuse angled.

Maths

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