Question

Let \( \alpha \) be the solution of the equation \[ 16 \sin^2 \theta + 16 \cos^2 \theta = 10, \quad \theta \in \left( 0, \frac{\pi}{4} \right) \] {If the shadow of a vertical pole is \( \frac{1}{\sqrt{3}} \) of its height, then the altitude of the sun is:}

• A. \( \alpha \)
• B. \( \frac{\alpha}{2} \)
• C. \( 2\alpha \)
• D. \( \frac{\alpha}{3} \)

Hint:

In problems involving trigonometric functions and shadows, recall that the tangent of the angle of elevation is the ratio of the height to the length of the shadow. Use this to find the altitude and check the periodic nature of angles when solving equations in given intervals.

Answer 1

The Correct Option is C

Given the equation: \[ 16 \sin^6\theta + 16 \cos^6\theta = 10 \] This can be rewritten using the identity for power of cosines and sines: \[ 16\sin^6\theta + 16(1-\sin^6\theta) = 10 \] Letting \( x = \sin^2\theta \), we have: \[ x^3 - 10x + 16 = 0 \Rightarrow x = 2, 8 \] Thus, \[ \sin^2\theta = 2, \quad \sin^2\theta = 8 \Rightarrow \sin^2\theta = \frac{1}{4} \left(\sqrt{3}\right)^2 \] Which gives: \[ \sin\theta = \frac{1}{2}\sqrt{3} \quad {thus, } \theta = \frac{\pi}{6} \] Consider the right triangle with altitude \( h \): \[ \tan\theta = \frac{h}{\frac{h}{\sqrt{3}}} = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3} { which is } 2\theta \] Thus, the correct answer is Option C.