Question

In a triangle ABC,$\frac{abc}{2∆}$ is equal to

• A. $(r_1+r_2)/(1+\cos C)$
• B. $(r_1+r_2)/(1+\sin C)$
• C. $(r_1+r_2)/(1+\tan C)$
• D. None of these

Answer 1

The correct option is A

$(r_1+r_2)/(1+\cos C)$

Explanation for the correct option:

Finding the value of $\frac{abc}{2∆}$:

We know that radius of the ex-circles:

$$\begin{gathered} r_1=\frac{∆}{(s–a)} \\ {r}_2=\frac{∆}{(s–b)} \\ \cos C/2=\sqrt{\left(\right(s(s-c)/ab)} \end{gathered}$$

Let's take option $A$, then

$\begin{array}{rcl}\frac{(r_1+r_2)}{(1+\cos C)}& =& [\frac{∆}{(s–a)}+\frac{∆}{(s–b)}]\left(\frac{1}{2}se{c}^2\frac{C}{2}\right)\\ & =& \left[\frac{∆(s–b+s–a)}{(s–a)(s–b)}\right]\left(\frac{1}{2}ab/s\left(s-c\right)\right)\\ & =& \frac{∆cab}{2s(s–a)(s–b)(s–c)}(\sin ce2s–a–b=c)\\ & =& \frac{∆abc}{2{∆}^2}\\ & =& \frac{abc}{2∆}\\ & =& \frac{abc}{2∆}\end{array}$

Hence, the correct option is (A).