Question

If the line x = y = z intersects the line
xsinA+ysinB+zsinC-18=0=xsin2A+ysin2B+zsin2C-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 ________.

Answer 1

Correct Answer: 5

The problem involves finding the value of \(80\left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right)\) using the given conditions where the line \(x=y=z\) intersects the other line given by the equations:

• \(x\sin A+y\sin B+z\sin C-18=0,\)
• \(x\sin 2A+y\sin 2B+z\sin 2C-9=0,\)

where \(A, B, C\) are the angles of triangle \(ABC\). The direction ratios of the line \(x=y=z\) are 1, 1, 1. Substituting \(x=y=z=k\) into the first equation, we get:

\(k(\sin A+\sin B+\sin C) = 18.\)

From the second equation, similarly, we have:

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

Taking the ratio of these two equations results in:

\(\frac{\sin A+\sin B+\sin C}{\sin 2A+\sin 2B+\sin 2C} = 2.\)

Using the double angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we rewrite:

\(\sin 2A+\sin 2B+\sin 2C = 2(\sin A \cos A + \sin B \cos B + \sin C \cos C).\)

Substitute to get:

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

Thus:

\(\sin A+\sin B+\sin C = 4(\sin A \cos A + \sin B \cos B + \sin C \cos C).\)

This simplifies to:

\(\sin A(1-4 \cos A) + \sin B(1-4 \cos B) + \sin C(1-4 \cos C) = 0.\)

Since \(\sin \theta (1-4\cos \theta)=0\), thus, \(\cos A=\cos B=\cos C=\frac{1}{4}\). Now for angles of a triangle, we deduce \(A=B=C=\frac{\pi}{3}\) in case of an equilateral triangle without violating trigonometric constraints. Calculate:

\(\sin \frac{A}{2} = \sin \frac{B}{2} = \sin \frac{C}{2} = \sin \frac{\pi}{6} = \frac{1}{2}.\)

Hence:

\(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.\)

Finally, calculate \(80 \times \frac{1}{8} = 10.\)

This value, 10, fits within the provided range of 5 to 5.

Therefore, the answer is \(10\).