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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

If α+ β+ γ=0 , where 0 < α ≤ (π)/(2), 0 < β ≤ (π)/(2) , 0 < γ ≤...

If cos⁡α+cos⁡β+cos⁡γ=0\cos \alpha+\cos \beta+\cos \gamma=0, where 0<α≤π2,0<β≤π20 < \alpha \leq \frac{\pi}{2}, 0 < \beta \leq \frac{\pi}{2}, 0<γ≤π20 < \gamma \leq \frac{\pi}{2}, then what is the value of sin⁡α+sin⁡β+sin⁡γ?\sin \alpha+\sin \beta+\sin \gamma ?

Options

  1. A00
  2. B33
  3. C522\frac{5 \sqrt{2}}{2}
  4. D322\frac{3 \sqrt{2}}{2}

Correct answer

B. 33

Step-by-step solution

cos⁡α+cos⁡β+cos⁡γ=0...(1)\cos \alpha+\cos \beta+\cos \gamma=0...(1) Given, 0<α≤π2,0<β≤π2,0<γ≤π20 < \alpha \leq \frac{\pi}{2}, 0 < \beta \leq \frac{\pi}{2}, 0 < \gamma \leq \frac{\pi}{2}. (1) is satisfied when α=π2,β=π2\alpha=\frac{\pi}{2}, \beta=\frac{\pi}{2} and γ=π2\gamma=\frac{\pi}{2}. ∴sin⁡α+sin⁡β+sin⁡γ=sin⁡π2+sin⁡π2+sin⁡π2=1+1+1=3\begin{aligned} & \therefore \sin \alpha+\sin \beta+\sin \gamma=\sin \frac{\pi}{2}+\sin \frac{\pi}{2}+\sin \frac{\pi}{2} \\ & =1+1+1=3 \end{aligned}
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