An isosceles triangle of wood of base 10 feet and height `(8)/(sqrt3)` feet is placed vertically with its base on the ground and vertex directly above. The triangle faces the sun whose altitude is `30^(@)`. Then, the tangent of the angle at the apex of the shadow is

Options

A. 80
B. `(80)/(39)`
C. `(89)/(2)`
D. `(80)/(217)`

Answer

To solve the problem, we need to find the tangent of the angle at the apex of the shadow of the isosceles triangle when the sun is at an altitude of \(30^\circ\). Here’s a step-by-step solution:

### Step 1: Understand the Triangle
We have an isosceles triangle with a base of 10 feet and a height of \(\frac{8}{\sqrt{3}}\) feet. The triangle is placed vertically, with the base on the ground.

### Step 2: Identify the Geometry
Let’s denote the vertices of the triangle as follows:
- Vertex A (apex of the triangle)
- Points B and C (the base points)

The height from A to the base BC is \(\frac{8}{\sqrt{3}}\) feet, and the base BC is 10 feet long. This means that the lengths of segments AB and AC are equal.

### Step 3: Calculate the Length of AB and AC
Using the Pythagorean theorem:
\[ AB = AC = \sqrt{\left(\frac{10}{2}\right)^2 + \left(\frac{8}{\sqrt{3}}\right)^2} \]
Calculating:
\[ AB = AC = \sqrt{5^2 + \left(\frac{8}{\sqrt{3}}\right)^2} = \sqrt{25 + \frac{64}{3}} = \sqrt{\frac{75 + 64}{3}} = \sqrt{\frac{139}{3}} = \frac{\sqrt{139}}{\sqrt{3}} \]

### Step 4: Determine the Length of the Shadow
When the sun is at an altitude of \(30^\circ\), the length of the shadow (DE) can be calculated using the cotangent of the angle:
\[ DE = \text{Height} \times \cot(30^\circ) = \frac{8}{\sqrt{3}} \times \sqrt{3} = 8 \text{ feet} \]

### Step 5: Calculate the Tangent of Angle Alpha
In triangle BDE, we can find \(\tan(\alpha)\):
\[ \tan(\alpha) = \frac{BD}{DE} \]
Where \(BD = \frac{10}{2} = 5\) feet and \(DE = 8\) feet:
\[ \tan(\alpha) = \frac{5}{8} \]

### Step 6: Find \(\tan(2\alpha)\)
Using the double angle formula for tangent:
\[ \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \]
Substituting \(\tan(\alpha) = \frac{5}{8}\):
\[ \tan(2\alpha) = \frac{2 \cdot \frac{5}{8}}{1 - \left(\frac{5}{8}\right)^2} = \frac{\frac{10}{8}}{1 - \frac{25}{64}} = \frac{\frac{10}{8}}{\frac{39}{64}} = \frac{10 \cdot 64}{8 \cdot 39} = \frac{80}{39} \]

### Final Answer
Thus, the tangent of the angle at the apex of the shadow is:
\[ \boxed{\frac{80}{39}} \]