If a flag-staff of 6 m height placed on the top of a tower throws shadow of 2√3 m along the ground, then what is the angle that the sun makes with the ground?
60°
Understanding the Problem: Flagstaff, Shadow, and Sun's Angle
The problem asks us to find the angle that the sun makes with the ground, given the height of a flag-staff placed on top of a tower and the length of the shadow it casts on the ground. This scenario can be modeled as a right-angled triangle where:
- The height of the flag-staff represents the side opposite the angle of elevation (the angle the sun's rays make with the ground).
- The length of the shadow represents the side adjacent to the angle of elevation along the ground.
- The line of sight from the tip of the flag-staff to the end of the shadow represents the hypotenuse.
We are given:
- Height of the flag-staff (Opposite side) = 6 m
- Length of the shadow (Adjacent side) = \(2\sqrt{3}\) m
We need to find the angle of elevation, let's call it \(\theta\).
Applying Trigonometry to Find the Angle
In a right-angled triangle, the tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side. This relationship is perfect for solving our problem involving height and shadow.
The formula is: $$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $$
Substituting the given values:
$$ \tan(\theta) = \frac{6 \text{ m}}{2\sqrt{3} \text{ m}} $$
Solving the Trigonometric Equation
Now, we need to simplify the expression and find the value of \(\theta\) for which the tangent is equal to the simplified value.
$$ \tan(\theta) = \frac{6}{2\sqrt{3}} $$
Simplify the fraction:
$$ \tan(\theta) = \frac{3}{\sqrt{3}} $$
To make the denominator rational, we can multiply the numerator and the denominator by \(\sqrt{3}\):
$$ \tan(\theta) = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} $$
$$ \tan(\theta) = \frac{3\sqrt{3}}{3} $$
$$ \tan(\theta) = \sqrt{3} $$
Now we need to find the angle \(\theta\) whose tangent is \(\sqrt{3}\). We recall the standard trigonometric values for common angles.
Finding the Angle of Elevation
We know that:
- \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)
- \(\tan(45^\circ) = 1\)
- \(\tan(60^\circ) = \sqrt{3}\)
Since we found that \(\tan(\theta) = \sqrt{3}\), the angle \(\theta\) must be \(60^\circ\).
Therefore, the angle that the sun makes with the ground is \(60^\circ\).
Conclusion
Based on our calculation, the angle of elevation of the sun is \(60^\circ\). This matches one of the given options.
Revision Table: Standard Trigonometric Ratios
It is helpful to remember the tangent values for common angles when solving such problems.
| Angle (\(\theta\)) | \(\tan(\theta)\) |
|---|---|
| \(0^\circ\) | 0 |
| \(30^\circ\) | \(\frac{1}{\sqrt{3}}\) |
| \(45^\circ\) | 1 |
| \(60^\circ\) | \(\sqrt{3}\) |
| \(90^\circ\) | Undefined |
Additional Information: Angle of Elevation
The angle of elevation is the angle formed by the horizontal line of sight and the line of sight upwards to an object. In this problem, the 'object' is the top of the flag-staff, and the horizontal line is the ground. The sun's rays are parallel, so the angle of elevation to the top of the flag-staff is the same as the angle the sun makes with the ground at that point.
Conversely, the angle of depression is the angle formed by the horizontal line of sight and the line of sight downwards to an object. Both angles are measured relative to a horizontal line.
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