JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2A + y \sin 2B + z \sin 2C - 9$ , 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 ________.

Choose the correct answer:

Explanation

Humein do equations mili thi:

$k(\sin A + \sin B + \sin C) = 18$
$k(\sin 2A + \sin 2B + \sin 2C) = 9$

Dono ko divide karne par:

\frac{\sin A + \sin B + \sin C}{\sin 2A + \sin 2B + \sin 2C} = \frac{18}{9} = 2

Ab triangle ki identities use karte hain:

$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Inhe equation mein rakhte hain:

\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{4 \sin A \sin B \sin C} = 2

Ab hum jaante hain ki $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$ hota hai. Toh denominator ko expand karte hain:

\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{4 (2 \sin \frac{A}{2} \cos \frac{A}{2}) (2 \sin \frac{B}{2} \cos \frac{B}{2}) (2 \sin \frac{C}{2} \cos \frac{C}{2})} = 2

Saari $\cos$ terms cancel ho jayengi:

\frac{4}{4 \cdot 8 \cdot (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})} = 2

\frac{1}{8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 2

Ab $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ ko upar le jayein aur 2 ko niche:

\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{8 \times 2} = \frac{1}{16}

Final Calculation:

Humein value nikalni hai $80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right)$ ki:

80 \times \frac{1}{16} = 5

Choose the correct answer:

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