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.

Solution

## 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, $$\cos C=\displaystyle \frac{a^2+b^2-c^2}{2ab}$$$$\cos C=\displaystyle \frac { (3x+4y)^{ 2 }+(4x+3y)^{ 2 }-(5x+5y)^{ 2 } }{ 2(3x+4y)(4x+3y) }$$$$\cos C=\displaystyle \frac { 9x^{ 2 }+16y^{ 2 }+24xy+16x^{ 2 }+9y^{ 2 }+24xy-25x^{ 2 }-25y^{ 2 }-50xy }{ 2(3x+4y)(4x+3y) }$$$$\cos C=\cfrac { -2xy }{ 2(3x+4y)(4x+3y) }$$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.Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More