In a triangle ABC,with usual notations,2 ac s | vedclass

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

Question

(C) We know that in any triangle $ABC$,$A+B+C = \pi$. Therefore,$A+C = \pi - B$. Substituting this into the expression: $2 ac \sin \left(\frac{A-B+C}{

Vedclass · Accepted Answer

$c^2+a^2-b^2$

Vedclass · Answer

(C) We know that in any triangle $ABC$,$A+B+C = \pi$. Therefore,$A+C = \pi - B$. Substituting this into the expression: $2 ac \sin \left(\frac{A-B+C}{2}\right) = 2 ac \sin \left(\frac{(A+C)-B}{2}\right)$ $= 2 ac \sin \left(\frac{(\pi-B)-B}{2}\right)$ $= 2 ac \sin \left(\frac{\pi-2B}{2}\right)$ $= 2 ac \sin \left(\frac{\pi}{2} - B\right)$ $= 2 ac \cos B$ Using the cosine rule,$\cos B = \frac{a^2+c^2-b^2}{2ac}$. Substituting this value: $= 2 ac \left(\frac{a^2+c^2-b^2}{2ac}\right)$ $= a^2+c^2-b^2$.

In a triangle $ABC$,with usual notations,$2 ac \sin \left(\frac{A-B+C}{2}\right)$ is equal to

Similar Questions

In $\triangle ABC$,$\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c} = $

In triangle $ABC$,if $a=13, b=8, c=7$,then $\cos(B+C) = $

If the angles of a triangle are in the ratio $1: 1: 4$,then the ratio of the perimeter of the triangle to its largest side is

In a triangle $ABC$,with usual notations,if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$,then $\cos A : \cos B : \cos C =$

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module