From the top of a hill 240 m high, the angles of depression of the top and bottom of a pole are 30° and 60°, respectively. The difference (in m) between the height of the pole and its distance from the hill is:
80 (2 - √3)
Understanding the Problem: Angles of Depression and Heights
The problem involves finding the difference between the height of a pole and its distance from a hill, given the height of the hill and the angles of depression from the top of the hill to the top and bottom of the pole. This is a classic application of trigonometry, specifically using angles of elevation and depression.
Setting up the Geometry
Let's visualize the scenario:
- Imagine the hill as a vertical line segment and the pole as another vertical line segment on the ground.
- The top of the hill is the point from where the angles of depression are measured.
- Angles of depression are measured downwards from a horizontal line of sight.
- The angle of depression to the bottom of the pole relates to the total height of the hill and the distance to the pole.
- The angle of depression to the top of the pole relates to the difference in height between the top of the hill and the top of the pole, and the distance to the pole.
We can form right-angled triangles using the hill, the ground, and the pole.
- Height of the hill, \(H = 240\) m.
- Let the height of the pole be \(h\) m.
- Let the distance of the pole from the base of the hill be \(d\) m.
The angle of depression from the top of the hill to the bottom of the pole is 60°. The alternate interior angle, which is the angle of elevation from the bottom of the pole (on the ground) to the top of the hill, is also 60°.
The angle of depression from the top of the hill to the top of the pole is 30°. The alternate interior angle, which is the angle of elevation from the top of the pole to the top of the hill, is also 30°.
Consider the point at the same height as the top of the pole, directly below the top of the hill. This point is \(H-h\) meters below the top of the hill. The horizontal distance from this point to the top of the pole is \(d\).
Calculations using Trigonometry
We can use the tangent function in the right-angled triangles formed.
Triangle 1: Using the angle of depression to the bottom of the pole (60°)
This triangle is formed by the hill's height (opposite side), the distance from the hill (adjacent side), and the line of sight to the bottom of the pole. The angle of elevation from the base of the pole to the top of the hill is 60°.
\[ \tan(60^\circ) = \frac{\text{Height of the hill}}{\text{Distance from the hill}} \] \[ \sqrt{3} = \frac{240}{d} \]Solving for \(d\):
\[ d = \frac{240}{\sqrt{3}} = \frac{240 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{240\sqrt{3}}{3} = 80\sqrt{3} \text{ m} \]Triangle 2: Using the angle of depression to the top of the pole (30°)
Consider a horizontal line from the top of the pole extending towards the hill. The vertical distance from the top of the hill to this horizontal line is \(H - h = 240 - h\). The horizontal distance is still \(d\). The angle of elevation from the top of the pole to the top of the hill is 30°.
\[ \tan(30^\circ) = \frac{\text{Vertical distance from hill top to pole top level}}{\text{Horizontal distance from hill to pole}} \] \[ \tan(30^\circ) = \frac{240 - h}{d} \] \[ \frac{1}{\sqrt{3}} = \frac{240 - h}{d} \]Substitute the value of \(d\) we found:
\[ \frac{1}{\sqrt{3}} = \frac{240 - h}{80\sqrt{3}} \]Multiply both sides by \(80\sqrt{3}\):
\[ 80\sqrt{3} \times \frac{1}{\sqrt{3}} = 240 - h \] \[ 80 = 240 - h \]Solving for \(h\):
\[ h = 240 - 80 = 160 \text{ m} \]Finding the Difference
We are asked to find the difference between the height of the pole (\(h\)) and its distance from the hill (\(d\)).
Difference \( = |h - d| \)
We have \(h = 160\) m and \(d = 80\sqrt{3}\) m.
Difference \( = |160 - 80\sqrt{3}| \)
Let's compare this with the options. It seems the option \(80(2 - \sqrt{3})\) is relevant. Let's check the value of \(160 - 80\sqrt{3}\).
\[ 160 - 80\sqrt{3} = 80(2 - \sqrt{3}) \]Since \(2 \approx 2.000\) and \(\sqrt{3} \approx 1.732\), \(2 - \sqrt{3}\) is positive. Thus, the difference is \(80(2 - \sqrt{3})\) m.
The final answer is \(80(2 - \sqrt{3})\) m.
Revision Table: Angles of Depression Problem
| Concept | Description | Formula Used |
|---|---|---|
| Angle of Depression | Angle below the horizontal from the observer to an object. Equal to the angle of elevation from the object to the observer. | N/A |
| Tangent Function (\(\tan\)) | Ratio of the length of the opposite side to the length of the adjacent side in a right-angled triangle. | \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \) |
| Special Angle Values | Specific values for trigonometric functions at common angles like 30° and 60°. | \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), \( \tan(60^\circ) = \sqrt{3} \) |
| Problem Solving Steps | Identify given information, draw diagram, relate angles to sides, set up equations, solve for unknowns, calculate required value. | N/A |
Additional Information: Heights, Distances, and Trigonometry
Trigonometry is extensively used in solving problems related to heights and distances. The angles of elevation and depression are key concepts. The angle of elevation is the angle measured upwards from a horizontal line to a point above the observer. The angle of depression is the angle measured downwards from a horizontal line to a point below the observer.
In these problems, it's crucial to draw a clear diagram representing the situation. Identify the right-angled triangles and the angles involved. The trigonometric ratios (sine, cosine, and tangent) relate the angles of a right-angled triangle to the lengths of its sides.
- Sine (\(\sin\)): Opposite / Hypotenuse
- Cosine (\(\cos\)): Adjacent / Hypotenuse
- Tangent (\(\tan\)): Opposite / Adjacent
For angles of depression, remember that the angle of depression from point A to point B is equal to the angle of elevation from point B to point A, assuming the points and observer positions allow for forming a horizontal line and alternate interior angles.
Carefully selecting the appropriate triangle and trigonometric ratio is essential for setting up the correct equations to solve for unknown heights or distances.
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