Question

In a triangle the sum of two sides is $x$ and their product is $y$ such that $(x+z)(x-z)=y$ where $z$ is the third side of the triangle.

Greatest angle of the triangle is :

• A. $90$ ${}^o$
• B. $110$ ${}^o$
• C. $120$ ${}^o$
• D. $135$ ${}^o$

Answer 1

Answer: (C)

$120$ ${}^o$

Let sides of the triangles are $a,b,c$
Given $a+b=x,ab=y,c=z$ (i)
Given $(x+z)(x-z)=y\Rightarrow x^2-z^2=y$ (ii)
Now using cosine rule, $\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{(a+b{)}^2-c^2-2ab}{2ab}$
$\Rightarrow \cos C=\frac{x^2-z^2-2y}{2y}=\frac{y-2y}{2y}=-\frac{1}{2}$ $\therefore \angle C={120}^0$ which is greatest angle.