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

The angles of a triangle are in the ratio $3: 5: 10$. Then the ratio of the smallest side to the greatest side is:

In a triangle $ABC$,with usual notations,if $3b = a + c$,then $\cot \frac{A}{2} \cdot \cot \frac{C}{2} = $

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

If the angles $A, B$ and $C$ of a triangle $ABC$ are in the ratio $2:3:7$ respectively,then the sides $a, b$ and $c$ are respectively in the ratio

In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module