CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image