Question

In a triangle if the sum of two sides is $x$ and the product is $y,(x\ge 2\sqrt{y})$ such that $(x+z)(x-z)=y$ where $z$ is the third side of the triangle.

On the basis of the above information, answer the following question:

Greatest angle of the triangle is

• A. ${105}^0$
• B. ${120}^0$
• C. ${135}^0$
• D. ${150}^0$

Answer 1

The correct option is B ${120}^0$

Let the sides of triangles are $a,bandc$

$\therefore b+c=x,bc=y$

$\because$ A.M $\ge$ G.M

$\therefore \frac{b+c}{2}\ge \sqrt{bc}$

$\Rightarrow \frac{x}{2}\ge \sqrt{y}$

$\Rightarrow x\ge 2\sqrt{y}$ and $z=a$

Since $(x+z)(x-z)=y$

Then $(b+c+a)(b+c-a)=bc$

Since $a+b+c=2s$ we have $b+c=2s-a$

$\Rightarrow 2s(2s-2a)=bc$

$\Rightarrow \sqrt{\frac{s(s-a)}{bc}}=\frac{1}{2}$

$\Rightarrow \cos \frac{A}{2}=\cos \frac{\pi }{3}$

$\therefore A=\pi -\frac{\pi }{3}=\frac{2\pi }{3}={120}^0$