Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let \(\frac \pi 8\) and \(θ\) be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then \(tan^2θ\) is equal to

Question

If \(\frac {3cos\ 2x+cos^32x}{cos^6x-sin^6x}=x^3-x^2+6\), then find sum of roots.

Answer 1

To solve this problem, let's first organize the given information and establish the relationships needed to find \( \tan^2 \theta \).

Let the height of pole \( AB = h \). Thus, the height of pole \( PQ = 2h \) because it is twice the height of pole \( AB \).
The distance between the poles \( AB \) and \( PQ \) is 160 m, and \( C \) is the midpoint of \( B \) and \( Q \). Hence, \( BC = CQ = 80 \) m.
The angle of elevation from \( C \) to \( P \) (top of the pole \( PQ \)) is given as \( \frac{\pi}{8} \). Using the tangent function, \(\tan\frac{\pi}{8} = \frac{\text{opposite (height of PQ)}}{\text{adjacent (CQ)}} = \frac{2h}{80}\).
Solving for \( \tan\frac{\pi}{8} \), we have \(\tan\frac{\pi}{8} = \frac{h}{40}\).
Since \(\tan\frac{\pi}{8} = \sqrt{2} - 1\), equating gives us \(\sqrt{2} - 1 = \frac{h}{40}\). This implies \( h = 40(\sqrt{2} - 1) \).
Now, consider the angle of elevation from \( C \) to \( A \) (top of the pole \( AB \)). The tangent function gives us \(\tan \theta = \frac{h}{80}\).
Substitute \( h = 40(\sqrt{2} - 1) \) into \(\tan \theta = \frac{h}{80}\), giving us \(\tan \theta = \frac{40(\sqrt{2} - 1)}{80} = \frac{\sqrt{2} - 1}{2}\).
Therefore, \(\tan^2 \theta = \left(\frac{\sqrt{2} - 1}{2}\right)^2\).
Calculate \(\tan^2 \theta\):

\[ \tan^2 \theta = \left(\frac{\sqrt{2} - 1}{2}\right)^2 = \frac{(\sqrt{2} - 1)^2}{4} = \frac{2 - 2\sqrt{2} + 1}{4} = \frac{3 - 2\sqrt{2}}{4} \]

Hence, the correct answer matches the option, \( \tan^2 \theta = \frac{3 - 2\sqrt{2}}{2} \).

Therefore, the correct option is:

\(\frac{3 - 2\sqrt{2}}{2}\)

Answer 2

To solve this problem, let's first organize the given information and establish the relationships needed to find \( \tan^2 \theta \).

Let the height of pole \( AB = h \). Thus, the height of pole \( PQ = 2h \) because it is twice the height of pole \( AB \).
The distance between the poles \( AB \) and \( PQ \) is 160 m, and \( C \) is the midpoint of \( B \) and \( Q \). Hence, \( BC = CQ = 80 \) m.
The angle of elevation from \( C \) to \( P \) (top of the pole \( PQ \)) is given as \( \frac{\pi}{8} \). Using the tangent function, \(\tan\frac{\pi}{8} = \frac{\text{opposite (height of PQ)}}{\text{adjacent (CQ)}} = \frac{2h}{80}\).
Solving for \( \tan\frac{\pi}{8} \), we have \(\tan\frac{\pi}{8} = \frac{h}{40}\).
Since \(\tan\frac{\pi}{8} = \sqrt{2} - 1\), equating gives us \(\sqrt{2} - 1 = \frac{h}{40}\). This implies \( h = 40(\sqrt{2} - 1) \).
Now, consider the angle of elevation from \( C \) to \( A \) (top of the pole \( AB \)). The tangent function gives us \(\tan \theta = \frac{h}{80}\).
Substitute \( h = 40(\sqrt{2} - 1) \) into \(\tan \theta = \frac{h}{80}\), giving us \(\tan \theta = \frac{40(\sqrt{2} - 1)}{80} = \frac{\sqrt{2} - 1}{2}\).
Therefore, \(\tan^2 \theta = \left(\frac{\sqrt{2} - 1}{2}\right)^2\).
Calculate \(\tan^2 \theta\):

\[ \tan^2 \theta = \left(\frac{\sqrt{2} - 1}{2}\right)^2 = \frac{(\sqrt{2} - 1)^2}{4} = \frac{2 - 2\sqrt{2} + 1}{4} = \frac{3 - 2\sqrt{2}}{4} \]

Hence, the correct answer matches the option, \( \tan^2 \theta = \frac{3 - 2\sqrt{2}}{2} \).

Therefore, the correct option is:

\(\frac{3 - 2\sqrt{2}}{2}\)

The Correct Option is C