Question

In a $\Delta ABC,\cot \frac{A}{2}=30,\cot \frac{B}{2}=50,\cot \frac{C}{2}=70$ then the sides $a,b,c$ in ascending order is:

• A. $a,b,c$
• B. $a,c,b$
• C. $b,a,c$
• D. $c,b,a$

Answer 1

The correct option is D $c,b,a$

As we know by using the half angle formula in sides form, The value of
$\Rightarrow \cot \left(\frac{A}{2}\right)=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\frac{s(s-a)}{\Delta }$
$\Rightarrow \Delta \cdot \cot \left(\frac{A}{2}\right)=s(s-a)$

Similarly, $\Delta \cdot \cot \left(\frac{B}{2}\right)=s(s-b)$

and, $\Delta \cdot \cot \left(\frac{C}{2}\right)=s(s-c)$

Dividing each term by $s$, then also the ratio remains the same. So, $(s-a),(s-b)$ and $(s-c)$ are in ratio of $3:5:7$

As we know $s=\frac{a+b+c}{2}$

So putting value of $s$ and replacing ratio by constant $k$, we get $b+c-a=6k$$\cdots \left(i\right)$

$a+c-b=10k$$\cdots \left(ii\right)$

$a+b-c=14k$$\cdots \left(iii\right)$

Adding equation$\left(i\right)$ and equation$\left(ii\right)$, we get $2c=16k$

$\therefore c=8k$

Similarly, adding equation$\left(ii\right)$and equation$\left(iii\right)$, we get $2a=24k$

$\therefore a=12k$

Similarly, adding equation $\left(i\right)$ and equation $\left(iii\right),$ we get $2b=20k$

$\therefore b=10k$

So, $a,b$ and $c$ are in the ratio of $12:10:8$ or

$6:5:4$
Hence we have $c,b,a$ is in ascending order.