Question

In a $\Delta ABC$$\sin A - \cos B = \cos C$, the angle $B$ is
$
  {\text{A}}{\text{. }}\dfrac{\pi }{2} \\
  {\text{B}}{\text{. }}\dfrac{\pi }{3} \\
  {\text{C}}{\text{. }}\dfrac{\pi }{4} \\
  {\text{D}}{\text{. }}\dfrac{\pi }{6} \\
 $

Answer 1

Hint: Here it is given these are angles of a triangle so we can apply a property that the sum of all angles of a triangle is ${180^0}$ and use formula to proceed further.

Complete step-by-step answer:

From given
$\sin A = \cos B + \cos C$

Now using formula $\left( {\sin A = 2\sin \dfrac{A}{2}.\cos \dfrac{A}{2}} \right)$ and $\left( {\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}.\cos \dfrac{{C - D}}{2}} \right)$ we get
$2\sin \dfrac{A}{2}.\cos \dfrac{A}{2} = 2\cos \left( {\dfrac{{B + C}}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right)$
We know the sum of all angles of a triangle is ${180^0}$. So,$A + B + C = {180^0}$$\left( {\therefore \dfrac{{B + C}}{2} = {{90}^0} - \dfrac{A}{2}} \right)$
Now our question becomes,
$2\sin \dfrac{A}{2}.\cos \dfrac{A}{2} = 2\cos \left( {{{90}^0} - \dfrac{A}{2}} \right)\cos \left( {\dfrac{{B - C}}{2}} \right)$
$\left( {\because \cos \left( {{{90}^0} - \dfrac{A}{2}} \right) = \sin \dfrac{A}{2}} \right)$ We get,
$\sin \dfrac{A}{2}.\cos \dfrac{A}{2} = \sin \dfrac{A}{2}.\cos \left( {\dfrac{{B - C}}{2}} \right)$$ \Rightarrow \sin \dfrac{A}{2}\left[ {\cos \dfrac{A}{2} - \cos \left( {\dfrac{{B - C}}{2}} \right)} \right] = 0$
$\because \sin \dfrac{A}{2} \ne 0$ because any angle of triangle can’t be equal to zero. So
$\cos \dfrac{A}{2} - \cos \left( {\dfrac{{B - C}}{2}} \right) = 0$
Now we know $\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right).\sin \left( {\dfrac{{B - A}}{2}} \right)$ using this formula we get,
$2\sin \left( {\dfrac{{A + B - C}}{4}} \right)\sin \left( {\dfrac{{B - C - A}}{4}} \right) = 0$. SO,
Either $\sin \left( {\dfrac{{A + B - C}}{4}} \right) = 0$ or $\sin \left( {\dfrac{{B - C - A}}{4}} \right) = 0$
We know if $\sin \theta = 0 \Rightarrow \theta = 0$.
So, $\dfrac{{A + B - C}}{4} = 0 \Rightarrow A + B = C$ and $\dfrac{{B - C - A}}{4} = 0 \Rightarrow A + C = B$ $ \ldots \left( 1 \right)$
We know: $A + B + C = {180^0}$ on the value of the equation $\left( 1 \right)$ we get.
$2B = {180^0} \Rightarrow B = {90^0}$

Hence option $A = \dfrac{\pi }{2}$ is the correct option.

Note: Whenever you get these types of questions the key concept of solving is you have to start from given and you have to use standard results to solve further so if you want to solve these questions you should have remembered all the results.