Question

In a triangle ABC, a, b, c are the lengths of its sides and A,B,C are the angles of triangle ABC. The correct relation is given by.
$(A){\text{ }}(b - c)\left( {\dfrac{{B - C}}{2}} \right) = {\text{a cos}}\dfrac{A}{2}$
$(B){\text{ }}(b - c){\text{cos}}\dfrac{A}{2} = {\text{a sin}}\left( {\dfrac{{B - C}}{2}} \right)$
$(C){\text{ }}(b - c){\text{sin }}\left( {\dfrac{{B - C}}{2}} \right) = {\text{a cos}}\dfrac{A}{2}$
$(D){\text{ }}(b - c){\text{sin }}\dfrac{A}{2} = 2{\text{ a sin}}\left( {\dfrac{{B + C}}{2}} \right)$

Answer 1

In the question we had given a triangle also lengths and angles of triangle $ABC$ are given. We have given four options and we have to check which one out of four options is correct. We will check it by using properties and identities of triangles where angles and sides (lengths) of triangles are given.

Complete step-by-step answer:
In the question, we had given a triangle $ABC$and also given that $a,{\text{ }}b{\text{ and }}c$ are the length of the sides of triangle $ABC$and $A,B,C$ are the angles of the triangle $ABC$Them we have to check out that which option out of four is correct I option $(ii)$we have to prove.
$(b - c){\text{ cos}}\dfrac{A}{2} = {\text{a sin }}\left( {\dfrac{{B - C}}{2}} \right)$
We know that in triangle$ABC$
$a = 2R{\text{ sin }}A,b = 2R{\text{ sin }}B{\text{ and }}C = 2R{\text{ sin }}C$
Where $A,B,C$are the angles of the triangle $ABC$
Therefore we are taking the term$\dfrac{{b - c}}{a}$
On substituting the values of $a,{\text{ }}b,{\text{ }}c$have we get
$\Rightarrow \dfrac{{b - c}}{a} - \dfrac{{2R{\text{ sin }}B - 2R{\text{ sin }}C}}{{2R{\text{ }}\operatorname{sin} A}}$
$2$R is common in all the terms on the right hand side.
$\Rightarrow \dfrac{{b - c}}{a} - \dfrac{{2R({\text{sin }}B - {\text{sin }}C)}}{{2R{\text{ }}\operatorname{sin} A}}$
$2$R gets cancel in numerator & denominator
$\Rightarrow \dfrac{{b - c}}{a} - \dfrac{{\operatorname{sin} B - \operatorname{sin} C}}{{\operatorname{sin} A}}$
Applying the formula in numerator which is
$\operatorname{sin} x - \operatorname{sin} y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)$
We get,
$\dfrac{{b - c}}{a} - \dfrac{{2\cos \left( {\dfrac{{B + C}}{2}} \right)\operatorname{sin} \left( {\dfrac{{B - C}}{2}} \right)}}{{\operatorname{sin} A}}$
In the denomination applying the identity
$\operatorname{sin} 2x = 2\sin x\cos Ax,$ we get
$\dfrac{{b - c}}{a} - \dfrac{{2\cos \left( {\dfrac{{B + C}}{2}} \right)\operatorname{sin} \left( {\dfrac{{B - C}}{2}} \right)}}{{\operatorname{sin} \dfrac{A}{2}\cos \dfrac{A}{2}}}{\text{ }}..............{\text{(1)}}$
since sum of all the angles of triangle is ${180^0}$ or $\pi$ , so up get
$A + B + C = \pi$
or $B + C = \pi - A$
and the term $\cos \left( {\dfrac{{B + C}}{2}} \right)$ becomes $\cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{2}} \right)$
since $\cos \left( {\dfrac{\pi }{2} - x} \right) = {\text{sin }}x$
So $\cos \left( {\dfrac{{B + C}}{2}} \right){\text{ becomes sin }}\dfrac{A}{2}$
So equation $(1)$ becomes
$\Rightarrow \dfrac{{b - c}}{a} - \dfrac{{2\sin \dfrac{A}{2}\operatorname{sin} \left( {\dfrac{{B - C}}{2}} \right)}}{{2\operatorname{sin} \dfrac{A}{2}{\text{ }}\cos {\text{ }}\dfrac{A}{2}}}{\text{ }}$
The term $2$ sin $\dfrac{A}{2}$ gets cancel in numerator and denominator
$\Rightarrow \dfrac{{b - c}}{a} = \dfrac{{\operatorname{sin} \left( {\dfrac{{B - C}}{2}} \right)}}{{\cos {\text{ }}\dfrac{A}{2}}}{\text{ }}$
On cross multiplying it, we get
$\Rightarrow (b - c)\cos \dfrac{A}{2} = {\text{a sin}}\left( {\dfrac{{B - C}}{2}} \right)$
Hence option $(ii)$ is correct.

Note: Sum of all the angles of triangle area ${180^0}$ which is known as angle sum property of the triangles. Also in triangle, the angle $A$ is made opposite to side a and similarly for other which means opposite two sides $a,b{\text{ and }}c$ angles made are $A,B{\text{ and }}C$ respectively.