JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the line \(x = y = z\) intersects the line \(x \sin A+ y\) \(\sin B + z \sin C -18=0= x \sin 2 A + y \sin 2 B + z\) \(\sin 2 C -9\), where \(A , B , C\) are the angles of a triangle \(ABC\), then \(80\left(\sin \frac{ A }{2} \sin \frac{ B }{2} \sin \frac{ C }{2}\right)\) is equal to \(..........\).
- A \(5\)
- B \(4\)
- C \(3\)
- D \(2\)
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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