In a triangle ABC, 2ac sin left(frac{A-B+C}{2 | vedclass

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

Question

(B) We know that in a $	riangle ABC$, $A+B+C = \pi$, so $A+C = \pi - B$.Substituting this into the expression, we get $\frac{A+C-B}{2} = \frac{\pi-B-

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that in a $	riangle ABC$, $A+B+C = \pi$, so $A+C = \pi - B$.Substituting this into the expression, we get $\frac{A+C-B}{2} = \frac{\pi-B-B}{2} = \frac{\pi}{2} - B$.Thus, $2ac \sin \left(\frac{A-B+C}{2}ight) = 2ac \sin \left(\frac{\pi}{2}-Bight)$.Using the identity $\sin \left(\frac{\pi}{2}-	hetaight) = \cos 	heta$, we get $2ac \cos B$.From the Law of Cosines, $\cos B = \frac{a^2+c^2-b^2}{2ac}$.Substituting this, we get $2ac \left(\frac{a^2+c^2-b^2}{2ac}ight) = a^2+c^2-b^2$.

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

Similar Questions

With usual notations,in a triangle $ABC$,if $\theta$ is any real number,then $a \cos (B-\theta) + b \cos (A+\theta)$ is equal to:

In any $\triangle ABC$,with usual notations,$c(a \cos B - b \cos A) =$

If $p_1, p_2, p_3$ are altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively and if $\Delta$ is the area of the triangle,$S$ is the semi-perimeter of the triangle,then find the value of $\frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}$.

With usual notations, in $\triangle ABC$, $(b - c)^2 \cos^2 \frac{A}{2} + (b + c)^2 \sin^2 \frac{A}{2} = $

In a $\triangle ABC$,if $a=5, b=6, c=7$,then the length of the median drawn from $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module