Which of the following is true in a triangle | vedclass

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

Question

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

Vedclass · Accepted Answer

$(b - c)\cos \frac{A}{2} = a\sin \frac{B - C}{2}$

Vedclass · Answer

(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.Thus,$a = k\sin A$,$b = k\sin B$,and $c = k\sin C$.Consider the expression $\frac{b - c}{a} = \frac{\sin B - \sin C}{\sin A}$.Using the sum-to-product formula,$\sin B - \sin C = 2\sin \frac{B - C}{2} \cos \frac{B + C}{2}$.Since $A + B + C = \pi$,we have $\frac{B + C}{2} = \frac{\pi}{2} - \frac{A}{2}$,so $\cos \frac{B + C}{2} = \sin \frac{A}{2}$.Also,$\sin A = 2\sin \frac{A}{2} \cos \frac{A}{2}$.Substituting these into the expression:$\frac{b - c}{a} = \frac{2\sin \frac{B - C}{2} \sin \frac{A}{2}}{2\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\sin \frac{B - C}{2}}{\cos \frac{A}{2}}$.Rearranging gives $(b - c)\cos \frac{A}{2} = a\sin \frac{B - C}{2}$.

Which of the following is true in a triangle $ABC$?

Similar Questions

In a triangle $ABC$, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

In a $\triangle ABC$,if $3a = b + c$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

In a $\triangle ABC$,if $a=6$,$b=5$,and $c=4$,then find the value of $\cos 2A$.

In $\Delta ABC$,$\frac{\sin(A - B)}{\sin(A + B)} = $

In a $\triangle ABC$,if $a \cos A = b \cos B$,where $a \neq b$,then $\triangle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module