Question

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

Answer 1

Hint: In this question, we will use the law of sines, which are, $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$ and $\sin B-\sin C=2\cos \left( \dfrac{B+C}{2} \right).\sin \left( \dfrac{B-C}{2} \right)$ to find the correct relation. We will also use $\sin \left( A \right)=2\sin \left( \dfrac{A}{2} \right).\cos \left( \dfrac{A}{2} \right)$. We can also write $\dfrac{b-c}{a}=\dfrac{2r\sin B-2r\sin C}{2r\sin A}$, using the sine law.

Complete step-by-step solution -
It is given in the question that in a triangle $ABC,a,b,c$ are the length of its sides and $A,B,C$ are the angles of the triangle $ABC$.

We know that the law of sines says that the ratio of sine of one angle to the opposite side is the same ratio as that of all the three angles, $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k$. From this law of sine, we can write,
$\begin{align}
  & \dfrac{\left( b-c \right)}{a}=\dfrac{k\left( \sin B \right)-k\left( \sin C \right)}{k\sin \left( A \right)} \\
 & \Rightarrow \dfrac{\left( b-c \right)}{a}=\dfrac{k\left[ \left( \sin B \right)-\left( \sin C \right) \right]}{k\sin \left( A \right)} \\
 & \Rightarrow \dfrac{\left( b-c \right)}{a}=\dfrac{\left( \sin B \right)-\left( \sin C \right)}{\sin \left( A \right)} \\
\end{align}$
We know that $\sin B-\sin C=2\cos \left( \dfrac{B+C}{2} \right).\sin \left( \dfrac{B-C}{2} \right)$ and $\sin \left( A \right)=2\sin \left( \dfrac{A}{2} \right).\cos \left( \dfrac{A}{2} \right)$, so by applying that in the above equation, we get,
$\dfrac{\left( b-c \right)}{a}=\dfrac{2\cos \left( \dfrac{B+C}{2} \right).\sin \left( \dfrac{B-C}{2} \right)}{2\sin \left( \dfrac{A}{2} \right).\cos \left( \dfrac{A}{2} \right)}\ldots \ldots \ldots \left( i \right)$
We know that $A+B+C=\pi $, as the sum of all the angles of a triangle is equal to $\pi $. So, we can write $B+C=\pi -A$. Therefore, the angle $\cos \left( \dfrac{B+C}{2} \right)$ as $\cos \left( \dfrac{\pi -A}{2} \right)=\cos \left( \dfrac{\pi }{2}-\dfrac{A}{2} \right)$. Then we can use $\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$, so we get, $\cos \left( \dfrac{\pi }{2}-\dfrac{A}{2} \right)=\sin \dfrac{A}{2}$. So, by substituting the value in the equation (i), we get,
$\dfrac{\left( b-c \right)}{a}=\dfrac{2\sin \dfrac{A}{2}.\sin \left( \dfrac{B-C}{2} \right)}{2\sin \dfrac{A}{2}.\cos \dfrac{A}{2}}$
By cancelling the similar terms from the numerator and the denominator, we get,
$\dfrac{\left( b-c \right)}{a}=\dfrac{\sin \left( \dfrac{B-C}{2} \right)}{\cos \dfrac{A}{2}}$
Now, by cross multiplying, we get,
$\left( b-c \right)\cos \left( \dfrac{A}{2} \right)=a\sin \left( \dfrac{B-C}{2} \right)$
Hence, the correct answer to the given question is option B, $\left( b-c \right)\cos \left( \dfrac{A}{2} \right)=a\sin \left( \dfrac{B-C}{2} \right)$.

Note: It becomes much easier to solve this question with the use of the trigonometric formulas like, $\sin A=2\sin \left( \dfrac{A}{2} \right).\cos \left( \dfrac{A}{2} \right)$ and $\sin \left( B-C \right)=2\sin \left( \dfrac{B-C}{2} \right).\cos \left( \dfrac{B+C}{2} \right)$. Most of the students get stuck while solving this type of questions as they do not remember the formula of the trigonometric functions, especially the trigonometric functions of sub-angles. Therefore, it is advisable to memorise all the formulas.