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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2022 · Mathematics · Trigonometric Ratios & Identities

In a triangle A B C , A- B- C=0 What is angle B equal to?

In a triangle ABCA B C, sin⁡A−cos⁡B−cos⁡C=0\sin A-\cos B-\cos C=0 What is angle BB equal to?

Options

  1. Aπ6\frac{\pi}{6}
  2. Bπ4\frac{\pi}{4}
  3. Cπ3\frac{\pi}{3}
  4. Dπ2\frac{\pi}{2}

Correct answer

D. π2\frac{\pi}{2}

Step-by-step solution

In △ABCA+B+C=πsin⁡A=cos⁡B+cos⁡C=2cos⁡B+C2⋅cos⁡B−C2\triangle \mathrm{ABC}A+B+C=\pi\sin A=\cos B+\cos C=2 \cos \frac{B+C}{2} \cdot \cos \frac{B-C}{2}⇒sin⁡A=2cos⁡(π2−A2)⋅cos⁡(B−C2)[∵B+C=π−A]\begin{aligned} \Rightarrow \quad \sin A=2 \cos \left(\frac{\pi}{2}-\frac{A}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right) & \\ & {[\because B+C=\pi-A] }\end{aligned}$\begin{aligned} & \Rightarrow \quad 2 \sin \frac{A}{2} \cdot \cos \frac{A}{2}=2 \sin \frac{A}{2} \cdot \cos \left(\frac{B-C}{2}\right) \\ & \Rightarrow \quad \cos \frac{A}{2}=\cos \left(\frac{B-C}{2}\right)\end{aligned}⇒A2=B−C2⇒A=B−C⇒A+C=B⇒π−B=B⇒B=π2\begin{aligned} & \Rightarrow \quad \frac{A}{2}=\frac{B-C}{2} \quad \Rightarrow \quad \mathrm{ A}=\mathrm{B}-\mathrm{C} \\ & \Rightarrow \quad \mathrm{A}+\mathrm{C}=\mathrm{B} \quad \Rightarrow \quad \pi-\mathrm{B}=\mathrm{B} \\ & \Rightarrow \quad B=\frac{\pi}{2}\end{aligned}$
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