Question

A right angled triangle has perimeter of length $7$ and hypotenuse of length $3$. If $\theta$ is the largest non-right angle in the triangle, then the value of $\cos \theta$ equals

• A. $\frac{\sqrt{6}-\sqrt{2}}{4}$
• B. $\frac{4+\sqrt{2}}{6}$
• C. $\frac{4-\sqrt{2}}{3}$
• D. $\frac{4-\sqrt{2}}{6}$

Answer 1

The correct option is D $\frac{4-\sqrt{2}}{6}$

Let ABC be a right angled triangle with a,b,c its sides such that $A=\frac{\pi }{2}$ and $a=3$
$\therefore b^2+c^2=a^2$
$\Rightarrow b^2+c^2=9$ ....... (1)
Perimeter$\left(p\right)=a+b+c$
$\Rightarrow 7=3+b+c$
$\Rightarrow b+c=4$ ........ (2)
Solving equation (1) & equation (2), we get
$b=\frac{4+\sqrt{2}}{2}$ and $c=\frac{4-\sqrt{2}}{2}$
Since $b>c$, therefore $B$ is the largest non-right angle.
In right angled triangle $ABC$
$\cos B=\frac{c}{a}=\frac{4-\sqrt{2}}{6}$

$\therefore \cos \theta =\frac{4-\sqrt{2}}{6}$