When the shadow of a pole 'h' metres high is `(sqrt(3)h)/3` metres, what is the angle of elevation of the sun at that time ?

Answer

To find the angle of elevation of the sun when the shadow of a pole of height 'h' meters is \((\sqrt{3}h)/3\) meters, we can use the concept of trigonometry, specifically the tangent function.

### Step-by-step Solution:

1. **Identify the components**:
- Let the height of the pole be \(AB = h\) meters (this is the perpendicular side).
- Let the length of the shadow be \(BC = \frac{\sqrt{3}h}{3}\) meters (this is the base).

2. **Set up the tangent function**:
- The tangent of the angle of elevation \(\theta\) can be expressed as:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \]
- Substituting the values we have:
\[ \tan(\theta) = \frac{h}{\frac{\sqrt{3}h}{3}} \]

3. **Simplify the expression**:
- To simplify, we can multiply by the reciprocal of the denominator:
\[ \tan(\theta) = h \cdot \frac{3}{\sqrt{3}h} \]
- The \(h\) in the numerator and denominator cancels out:
\[ \tan(\theta) = \frac{3}{\sqrt{3}} \]

4. **Further simplify**:
- We can simplify \(\frac{3}{\sqrt{3}}\) as follows:
\[ \tan(\theta) = \frac{3 \cdot \sqrt{3}}{3} = \sqrt{3} \]

5. **Find the angle**:
- We know that \(\tan(60^\circ) = \sqrt{3}\). Therefore, we can conclude:
\[ \theta = 60^\circ \]

6. **Final answer**:
- The angle of elevation of the sun at that time is \(60^\circ\).