Question

In $ABC$, if $\cot \frac{A}{2}$ : $\cot \frac{B}{2}$ : $\cot \frac{C}{2}=3:7:9$, then $a:b:c=$

• A. $7:9:11$
• B. $14:11:6$
• C. $7:19:25$
• D. $8:6:5$

Answer 1

The correct option is D $8:6:5$

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

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

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

Since we know that, $\cot \left(\frac{A}{2}\right)$ $,$ $\cot \left(\frac{B}{2}\right)$ and $\cot \left(\frac{C}{2}\right)$ are in ratio, so if we multiply each of them with $△$, the ratio remains the same.

So, $△$$\cot \left(\frac{A}{2}\right)$ $,$ $△$$\cot \left(\frac{B}{2}\right)$ and $△$$\cot \left(\frac{C}{2}\right)$ are in ratio

$△$$\cot \left(\frac{A}{2}\right)$ $=$ $s(s-a)$

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

Similarly, $△$$\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 $3:7:9$

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

So putting value of $s$ and replacing ration by constant $k$, we get

$b+c-a=6k$

$a+c-b=14k$

$a+b-c=18k$

Adding equation (1) and equation (2), we get

$2c=20k$

$\therefore$ $c=10k$

Similarly, adding equation (2) and equation (3), we get

$2a=32k$

$\therefore$ $a=16k$

Similarly, adding equation (1) and equation (3), we get

$2b=24k$

$\therefore$ $b=12k$

So, $a,b$ and $c$ are in the ratio of $16:12:10$, on diving the ratio by 2 we get, $8:6:5$.

Hence, option D is the correct answer.