In triangle ABC with usual notations, if 1 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $	an(\frac{A}{2}) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an(\frac{B}{2}) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Multiplying these,

Vedclass · Accepted Answer

$a + b$

Vedclass · Answer

(C) We know that $	an(\frac{A}{2}) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an(\frac{B}{2}) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Multiplying these, we get $	an(\frac{A}{2}) 	an(\frac{B}{2}) = \sqrt{\frac{(s-b)(s-c)(s-a)(s-c)}{s(s-a)s(s-b)}} = \sqrt{\frac{(s-c)^2}{s^2}} = \frac{s-c}{s}$.Substituting this into the given expression: $1 + \frac{s-c}{s} = \frac{s + s - c}{s} = \frac{2s - c}{s}$.Since $2s = a + b + c$, we have $\frac{a + b + c - c}{s} = \frac{a + b}{s}$.Comparing this with $\frac{k}{s}$, we get $k = a + b$.

In $\triangle ABC$ with usual notations, if $1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) = \frac{k}{s}$ (where $s$ is the semi-perimeter), then the value of $k$ is...

Similar Questions

If in a triangle $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$,then the triangle is

What is the value of $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$?

In a $\triangle ABC$,the altitudes from $B$ and $C$ to the opposite sides are not shorter than their respective opposite sides. Then,one of the angles of $\triangle ABC$ is $........^{\circ}$

In a $\triangle ABC$,if $a^4+b^4+c^4=2b^2c^2+2a^2b^2$,then $B=$

In a $\triangle ABC$,if $a+c=5b$,then $\cot \frac{A}{2} \cot \frac{C}{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module