Question

A triangle has base $10cm$ and base angles are ${50}^0,{70}^0.$ If the perimeter of $\Delta$ is $x+y\cos z^0$ where $z^0\in (0,{90}^0),$ then $x+y+z=$

• A. $60$
• B. $55$
• C. $50$
• D. $40$

Answer 1

The correct option is D $40$

Given: Base angles$={50}^0,{70}^0$ and $a=10$
We know, $b+c=4R\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)$
$=4R\frac{\sqrt{3}}{2}\times \cos {10}^0=2R\sqrt{3}\cos {10}^0$
We know $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
Also, $A={60}^{\circ }$
Hence, $\frac{2a}{\sqrt{3}}=2R\Rightarrow 2R\sqrt{3}=20$
$\therefore b+c=20\cos {10}^0$
So, we have perimeter $=a+b+c=10+20\cos {10}^0=x+y\cos z^0$
$\therefore x=10,y=20,z=10$
$\Rightarrow x+y+z=40$