A pole 6 m high casts a shadow 2`sqrt 3` m long on the ground, then find the angle of elevation of the sun.

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `90^(@)`

Explanation

To find the angle of elevation of the sun given a pole of height 6 m and a shadow of length \(2\sqrt{3}\) m, we can follow these steps:

### Step 1: Understand the problem
We have a vertical pole (height = 6 m) and a horizontal shadow (length = \(2\sqrt{3}\) m). We need to find the angle of elevation of the sun, which is the angle formed between the line from the top of the pole to the tip of the shadow and the horizontal ground.

### Step 2: Draw the triangle
Let's denote:
- The height of the pole as \(AB = 6\) m (perpendicular).
- The length of the shadow as \(BC = 2\sqrt{3}\) m (base).
- The angle of elevation as \(\theta\).

This forms a right triangle \(ABC\) where:
- \(AB\) is the opposite side (height of the pole),
- \(BC\) is the adjacent side (length of the shadow),
- \(AC\) is the hypotenuse (the line from the top of the pole to the tip of the shadow).

### Step 3: Use the tangent function
The tangent of the angle \(\theta\) is defined as the ratio of the opposite side to the adjacent side:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{6}{2\sqrt{3}} \]

### Step 4: Simplify the expression
Now we simplify the expression:
\[ \tan(\theta) = \frac{6}{2\sqrt{3}} = \frac{6 \div 2}{\sqrt{3}} = \frac{3}{\sqrt{3}} \]
To rationalize the denominator, we multiply the numerator and denominator by \(\sqrt{3}\):
\[ \tan(\theta) = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

### Step 5: Find the angle
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we can conclude:
\[ \theta = 60^\circ \]

### Final Answer
The angle of elevation of the sun is \(60^\circ\).

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