Question

In a triangle ABC, Sin A –Cos B= Cos C then what is B equal to?
A. $ \pi $
B. $ \dfrac{\pi }{3} $
C. $ \dfrac{\pi }{2} $
D. $ \dfrac{\pi }{4} $

Answer 1

Hint: We know that the sum of inner angles in a triangle is 180 by using this we can substitute A (because we need to find B so we will not substitute B). Put Cos C and Cos B in LHS then use Cos C + Cos D formula.

Complete step-by-step answer:
It is given that,
 $ \operatorname{Sin}A-\operatorname{Cos}B=\operatorname{Cos}C $
Add Cos B to both sides,
 $ \operatorname{Sin}A=\operatorname{Cos}B+\operatorname{Cos}C $
Now we can use $ \operatorname{Cos}x+\operatorname{Cos}y=2.\operatorname{Cos}\dfrac{x+y}{2}.\operatorname{Cos}\dfrac{x-y}{2} $
Apply it on LHS,
 $ \operatorname{Sin}A=2.\operatorname{Cos}\dfrac{B+C}{2}.\operatorname{Cos}\dfrac{B-C}{2} $ …… (1)
We know that in a triangle,
  $ \begin{align}
  & A+B+C={{180}^{\circ }} \\
 & B+C={{180}^{\circ }}-A \\
 & \dfrac{B+C}{2}={{90}^{\circ }}-\dfrac{A}{2} \\
\end{align} $
Substitute this value in equation (1).
 $ \operatorname{Sin}A=2.\operatorname{Cos}\left( 90-\dfrac{A}{2} \right).\operatorname{Cos}\dfrac{(B-C)}{2} $
 $ \operatorname{Sin}A=2.\operatorname{Sin}\dfrac{A}{2}.\operatorname{Cos}\dfrac{(B-C)}{2} $ $ (\because \operatorname{Cos}(90-x)=\operatorname{Sin}x) $
 $ 2.\operatorname{Sin}\dfrac{A}{2}.\operatorname{Cos}\dfrac{A}{2}=2.\operatorname{Sin}\dfrac{A}{2}.\operatorname{Cos}\dfrac{(B-C)}{2} $ $ (\because \operatorname{Sin}2x=2.\operatorname{Sin}x.\operatorname{Cos}x) $
We can cancel 2.SinA/2 from both sides.
 $ \begin{align}
  & \operatorname{Cos}\dfrac{A}{2}=\operatorname{Cos}\dfrac{(B-C)}{2} \\
 & \dfrac{A}{2}=\dfrac{(B-C)}{2} \\
 & A=B-C \\
 & A+C=B \\
 & {{180}^{\circ }}-B=B \\
 & B={{90}^{\circ }} \\
\end{align} $

Note:- All the angles are in degree form. We can also verify options by putting each value given in the options. To avoid mistakes, recall the behavior of trigonometry ratios in all coordinates.