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

જો રેખા \(x=y=z\) એ રેખા  \(x \sin A+y \sin B+z \sin C-18=0=x \sin 2 A+y \sin 2 B+z \sin 2 C-9\) ને છેદે,જ્યાં \(A, B, C\) એ ત્રિકોણ \(A B C\), ના ખૂણાઓ છે, તો \(80\left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)=.........\)

  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

\(\sin A +\sin B+\sin C=\frac{18}{x}\) \(\sin 2 A+\sin 2 B+\sin 2 C=\frac{9}{x}\) \(\therefore \sin A +\sin B+\sin C=2(\sin 2 A+\sin 2 B+\sin 2 C)\) \(4 \cos A / 2 \cos B / 2 \cos C / 2=2(4 \sin A \sin B \sin C)\) \(16 \sin A / 2 \sin B / 2 \sin C / 2=1\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app