In a triangle ABC,(b+c) sin frac{A}{2} = | vedclass

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

Question

(B) Using the Sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$b = k \sin B$ and $c = k \sin C$. Substituting the

Vedclass · Accepted Answer

$a \cos \left(\frac{B-C}{2}\right)$

Vedclass · Answer

(B) Using the Sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$b = k \sin B$ and $c = k \sin C$. Substituting these into the expression: $(b+c) \sin \frac{A}{2} = k(\sin B + \sin C) \sin \frac{A}{2}$. Using the sum-to-product formula $\sin B + \sin C = 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}$: $= k \left[ 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} \right] \sin \frac{A}{2}$. Since $A+B+C = \pi$,we have $\frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2}$,so $\sin \frac{B+C}{2} = \cos \frac{A}{2}$. $= k \left[ 2 \cos \frac{A}{2} \cos \frac{B-C}{2} \right] \sin \frac{A}{2}$. $= k \left[ 2 \sin \frac{A}{2} \cos \frac{A}{2} \right] \cos \frac{B-C}{2}$. $= k \sin A \cos \frac{B-C}{2}$. Since $a = k \sin A$,the expression becomes $a \cos \frac{B-C}{2}$.

In a triangle $ABC$,$(b+c) \sin \frac{A}{2} =$

Similar Questions

In triangle $ABC$,if $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$,then $\frac{a^2-b^2}{c^2}=$

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^o$. If the third side is $3$,the remaining fourth side is

In a triangle $ABC$,with usual notations,if $\tan \left(\frac{A}{2}\right) = \frac{5}{6}$ and $\tan \left(\frac{C}{2}\right) = \frac{2}{5}$,then:

In $\triangle ABC$,if $a, b, c$ are its sides and $\angle C = 60^{\circ}$,find the value of $\frac{a}{b+c} + \frac{b}{c+a}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module