Let `alpha` be the solution of `16^(sin^2 theta)+ 16^(cos^2 theta)=10` in `(0,pi//4)` . If the shadow of a vertical pole is `1/sqrt3` of its height , then the altitude of the sun is

Options

A. `alpha`
B. `alpha/2`
C. `2alpha`
D. `alpha/3`

Explanation

To solve the problem step by step, we will break down the given information and equations systematically.

### Step 1: Solve the equation
We start with the equation given in the problem:

\[ 16^{\sin^2 \theta} + 16^{\cos^2 \theta} = 10 \]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can rewrite \( \cos^2 \theta \) as \( 1 - \sin^2 \theta \):

\[ 16^{\sin^2 \theta} + 16^{1 - \sin^2 \theta} = 10 \]

This simplifies to:

\[ 16^{\sin^2 \theta} + \frac{16}{16^{\sin^2 \theta}} = 10 \]

Let \( x = 16^{\sin^2 \theta} \). Then the equation becomes:

\[ x + \frac{16}{x} = 10 \]

### Step 2: Multiply through by \( x \)
To eliminate the fraction, multiply the entire equation by \( x \):

\[ x^2 + 16 = 10x \]

Rearranging gives us a standard quadratic equation:

\[ x^2 - 10x + 16 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = -10, c = 16 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-10)^2 - 4 \cdot 1 \cdot 16 = 100 - 64 = 36 \]

Now substituting back into the formula:

\[ x = \frac{10 \pm \sqrt{36}}{2 \cdot 1} = \frac{10 \pm 6}{2} \]

This gives us two solutions:

\[ x_1 = \frac{16}{2} = 8, \quad x_2 = \frac{4}{2} = 2 \]

### Step 4: Relate back to \( \sin^2 \theta \)
Recall that \( x = 16^{\sin^2 \theta} \). Thus, we have:

1. \( 16^{\sin^2 \theta} = 8 \) implies \( \sin^2 \theta = \frac{3}{4} \)
2. \( 16^{\sin^2 \theta} = 2 \) implies \( \sin^2 \theta = \frac{1}{4} \)

Calculating the angles:

- For \( \sin^2 \theta = \frac{3}{4} \), \( \sin \theta = \frac{\sqrt{3}}{2} \) gives \( \theta = \frac{\pi}{3} \) (not in \( (0, \frac{\pi}{4}) \)).
- For \( \sin^2 \theta = \frac{1}{4} \), \( \sin \theta = \frac{1}{2} \) gives \( \theta = \frac{\pi}{6} \) (in \( (0, \frac{\pi}{4}) \)).

Thus, we have:

\[ \alpha = \frac{\pi}{6} \]

### Step 5: Determine the altitude of the sun
Given that the shadow of a vertical pole is \( \frac{1}{\sqrt{3}} \) times its height, we can denote the height of the pole as \( h \) and the length of the shadow as \( \frac{h}{\sqrt{3}} \).

Using the tangent function:

\[ \tan \theta = \frac{h}{\frac{h}{\sqrt{3}}} = \sqrt{3} \]

This implies:

\[ \theta = \frac{\pi}{3} \]

### Step 6: Relate \( \theta \) and \( \alpha \)
From the previous steps, we have:

\[ \theta = 2\alpha \]

Substituting \( \alpha = \frac{\pi}{6} \):

\[ \theta = 2 \times \frac{\pi}{6} = \frac{\pi}{3} \]

### Conclusion
Thus, the altitude of the sun in terms of \( \alpha \) is:

\[ \theta = 2\alpha \]