If the line x = y = z intersects the line def | vedclass

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(A) Let the point of intersection be $(k, k, k)$.Since this point lies on the planes $x \sin A + y \sin B + z \sin C = 18$ and $x \sin 2A + y \sin 2B

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the point of intersection be $(k, k, k)$.Since this point lies on the planes $x \sin A + y \sin B + z \sin C = 18$ and $x \sin 2A + y \sin 2B + z \sin 2C = 9$,we have:$k(\sin A + \sin B + \sin C) = 18 \implies \sin A + \sin B + \sin C = \frac{18}{k}$$k(\sin 2A + \sin 2B + \sin 2C) = 9 \implies \sin 2A + \sin 2B + \sin 2C = \frac{9}{k}$Dividing the two equations,we get $\frac{\sin A + \sin B + \sin C}{\sin 2A + \sin 2B + \sin 2C} = \frac{18}{9} = 2$.Thus,$\sin A + \sin B + \sin C = 2(\sin 2A + \sin 2B + \sin 2C)$.Using the identities $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ and $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C = 4(2 \sin \frac{A}{2} \cos \frac{A}{2})(2 \sin \frac{B}{2} \cos \frac{B}{2})(2 \sin \frac{C}{2} \cos \frac{C}{2}) = 32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.Substituting these into the equation: $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = 2(32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2})$.Canceling $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ (which is non-zero),we get $1 = 16 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.Therefore,$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$.Finally,$80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} ight) = 80 	imes \frac{1}{16} = 5$.

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