Question

In a triangle $ABC$, if $2R+r=r_1$, then?

• A. $\angle A=\frac{\pi }{2}$
• B. $\angle B=\frac{\pi }{2}$
• C. $\angle C=\frac{\pi }{2}$
• D. $\angle A=\frac{\pi }{3}$

Answer 1

The correct option is A $\angle A=\frac{\pi }{2}$

In $\Delta ABC$

for $\angle A$ and side $a$

We know $R=\frac{a}{2\sin A}$

$r=(s-a)\tan \frac{A}{2}$

And $r_1=\left(s\right)\tan \frac{A}{2}$

Now put the all above value in given equation $2R+r=r_1$

$L.H.S.=\frac{a}{\sin A}+(s-a)\tan \frac{A}{2}$

$R.H.S=\left(s\right)\tan \frac{A}{2}$

Now equate $L.H.S.$ and $R.H.S.$

$\frac{a}{\sin A}+(s-a)\tan \frac{A}{2}$ =$\left(s\right)\tan \frac{A}{2}$

If we see carefully, only $A=\frac{\pi }{2}$ satisfies the above equation.

Hence, $A=\frac{\pi }{2}$