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JEE Mains · Maths · STD 12 - 11. three dimension geometry

यदि रेखा \(x=y=z\) रेखा \(x \sin A+y \sin B+\) \(\mathrm{z} \sin \mathrm{C}-18=0=\mathrm{x} \sin 2 \mathrm{~A}+\mathrm{y} \sin 2 \mathrm{~B}+\mathrm{z} \sin 2 \mathrm{C}-9\), को काटती है, जहाँ \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) एक त्रिभुज \(\mathrm{ABC}\) का कोण हैं, तो \(80\left(\sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}\right)\) बराबर है :

  1. A \(5\)
  2. B \(4\)
  3. C \(3\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(5\)

Step-by-step Solution

Detailed explanation

\(\sin A +\sin B+\sin C=\frac{18}{x}\) \(\sin 2 A+\sin 2 B+\sin 2 C=\frac{9}{x}\) \(\therefore \sin A +\sin B+\sin C=2(\sin 2 A+\sin 2 B+\sin 2 C)\) \(4 \cos A / 2 \cos B / 2 \cos C / 2=2(4 \sin A \sin B \sin C)\) \(16 \sin A / 2 \sin B / 2 \sin C / 2=1\)…
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