Q218 / 270 · NDA 2014 (Phase 2) · Mathematics

In a triangle $A B C, \sin A-\cos B=\cos C$, then what is $B$ euqal to?

  • A
    $\pi$
  • B
    $\frac{\pi}{3}$
  • Correct answer
    C
    $\frac{\pi}{2}$
  • D
    $\frac{\pi}{4}$

Solution

$\begin{aligned} & \because \quad \sin A-\cos B=\cos C \text { (given) } \\ & \Rightarrow \quad \cos B+\cos C=\sin A \\ & \Rightarrow \quad 2 \cos \frac{B+C}{2} \cdot \cos \frac{B-C}{2}=\sin A \\ & \Rightarrow \quad 2 \sin \frac{A}{2} \cdot \cos \frac{(B-C)}{2}=2 \sin \frac{A}{2} \cdot \cos \frac{A}{2}\end{aligned}$ $\Rightarrow \quad \cos \frac{(\mathrm{B}-\mathrm{C})}{2}=\cos \frac{\mathrm{A}}{2}$ $\Rightarrow \quad \frac{\mathrm{B}-\mathrm{C}}{2}=\frac{\mathrm{A}}{2}$ $\Rightarrow \quad \mathrm{B}-\mathrm{C}=\mathrm{A}$ $\Rightarrow \quad B=A+C$ $\because \quad A+B+C=180^{\circ}$ $\Rightarrow \quad B+B=180^{\circ}$ $\Rightarrow \quad 2 \mathrm{~B}=180^{\circ}$ $\Rightarrow \quad \mathrm{B}=90^{\circ}=\frac{\pi}{2}$

Hence, the correct answer is option: 'C'.

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