The angle of elevation of the top of a hill at the foot of the tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height of the hill?
150 m
This problem involves trigonometry, specifically using angles of elevation to find unknown heights. We have a tower and a hill, and we are given the heights of one (the tower) and angles of elevation between the top of one structure and the foot of the other.
The angle of elevation is the angle between the horizontal line from an observer's eye to an object, when the object is above the horizontal line. In this case, the horizontal line is the ground level between the tower and the hill.
We are given the height of the tower is 50 m. We need to find the height of the hill.
Let's imagine the situation forms two right-angled triangles. Let:
From the problem description, we can form two right-angled triangles:
We can use the tangent function, which relates the opposite side and the adjacent side in a right-angled triangle: \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\).
From Triangle 1 (using the tower):
\(\tan(30^\circ) = \frac{\text{Height of Tower}}{\text{Distance Between Bases}}\)
\(\tan(30^\circ) = \frac{50}{D}\)
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). So,
\(\frac{1}{\sqrt{3}} = \frac{50}{D}\)
Solving for \(D\):
\(D = 50 \times \sqrt{3}\) m
From Triangle 2 (using the hill):
\(\tan(60^\circ) = \frac{\text{Height of Hill}}{\text{Distance Between Bases}}\)
\(\tan(60^\circ) = \frac{H}{D}\)
We know that \(\tan(60^\circ) = \sqrt{3}\). So,
\(\sqrt{3} = \frac{H}{D}\)
Now, substitute the value of \(D\) we found in Step 1:
\(\sqrt{3} = \frac{H}{50\sqrt{3}}\)
Solving for \(H\):
\(H = \sqrt{3} \times (50\sqrt{3})\)
\(H = 50 \times (\sqrt{3} \times \sqrt{3})\)
\(H = 50 \times 3\)
\(H = 150\) m
So, the height of the hill is 150 m.
We used the given angles of elevation and the height of the tower to first find the horizontal distance between the two structures, and then used that distance along with the other angle of elevation to find the height of the hill.
| Trigonometric Ratio | Definition | Common Values |
|---|---|---|
| Sine (\(\sin \theta\)) | Opposite / Hypotenuse | \(\sin 30^\circ = 1/2\), \(\sin 60^\circ = \sqrt{3}/2\) |
| Cosine (\(\cos \theta\)) | Adjacent / Hypotenuse | \(\cos 30^\circ = \sqrt{3}/2\), \(\cos 60^\circ = 1/2\) |
| Tangent (\(\tan \theta\)) | Opposite / Adjacent | \(\tan 30^\circ = 1/\sqrt{3}\), \(\tan 60^\circ = \sqrt{3}\) |
Angles of elevation and depression are widely used in various fields:
These angles are crucial tools for indirect measurement using the principles of trigonometry.
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