Question

If $x,y>0$,then prove that the triangle whose sides are given by $3x+4y,4x+3y$ and $5x+5y$ units is obtused angled.

Answer 1

Given , $x,y>0$

So, $3x+4y,4x+3y,5x+5y$ all are greater than 0.

Let $a=3x+4y,b=4x+3y,c=5x+5y$ be the sides of the triangle .

Consider, $\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\cos C=\frac{(3x+4y{)}^2+(4x+3y{)}^2-(5x+5y{)}^2}{2(3x+4y)(4x+3y)}$

$\cos C=\frac{9x^2+16y^2+24xy+16x^2+9y^2+24xy-25x^2-25y^2-50xy}{2(3x+4y)(4x+3y)}$

$\cos C=\frac{-2xy}{2(3x+4y)(4x+3y)}$

Since, $x,y>0$

$\Rightarrow xy>0$

$\Rightarrow \cos C<0$ ($3x+4y>0,4x+3y>0$)

Since, $\cos C$ is negative in second and third quadrant . Since, C is angle of triangle , so it should be less than ${180}^0$

Hence, C lies in second quadrant i.e its measure is between ${90}^0$ and ${180}^0$

Hence, angle C is obtuse angle .

Hence, the triangle is obtuse angled triangle.