From the top of a platform 5 m high, the angle of elevation of a tower was 30°. If the tower was 45 m high, how far away from the tower was the platform positioned?
40√3m
This problem involves trigonometry, specifically the concept of the angle of elevation. We are given the heights of a platform and a tower, and the angle of elevation from the top of the platform to the top of the tower. We need to find the horizontal distance between the platform and the tower.
Let's visualize the scenario. We have two vertical lines representing the platform and the tower, standing on a horizontal ground. The observer is at the top of the platform. The line of sight goes from the top of the platform to the top of the tower. The angle of elevation is formed between the horizontal line from the top of the platform and the line of sight upwards to the top of the tower.
We are looking for the horizontal distance between the platform and the tower.
Draw a horizontal line from the top of the platform parallel to the ground, extending towards the tower. This line, the vertical segment from this line up to the top of the tower, and the line of sight form a right-angled triangle. The right angle is where the horizontal line meets the vertical line representing the tower.
The top of the platform is 5 m above the ground. The top of the tower is 45 m above the ground. The vertical height difference relevant to the triangle is the height of the tower above the level of the top of the platform.
Vertical height difference = Height of tower - Height of platform
Vertical height difference = \(45 \text{ m} - 5 \text{ m} = 40 \text{ m}\)
This 40 m is the length of the side opposite to the 30° angle in our right-angled triangle.
In the right-angled triangle, we know:
The trigonometric function that relates the opposite side and the adjacent side to an angle is the tangent function:
\(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)
Substituting the values:
\(\tan(30^\circ) = \frac{40 \text{ m}}{\text{Horizontal Distance}}\)
We know the value of \(\tan(30^\circ)\):
\(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)
So, the equation becomes:
\(\frac{1}{\sqrt{3}} = \frac{40}{\text{Horizontal Distance}}\)
To find the Horizontal Distance, we can rearrange the equation:
\(\text{Horizontal Distance} = 40 \times \sqrt{3}\)
\(\text{Horizontal Distance} = 40\sqrt{3} \text{ m}\)
The horizontal distance away from the tower that the platform was positioned is \(40\sqrt{3}\) m.
| Quantity | Value |
|---|---|
| Platform Height | 5 m |
| Tower Height | 45 m |
| Angle of Elevation | 30° |
| Vertical Difference (Opposite) | 40 m |
| Horizontal Distance (Adjacent) | ? |
| Trigonometric Relation | \(\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}}\) |
| Result | \(40\sqrt{3} \text{ m}\) |
| Term | Definition | Trig Ratio |
|---|---|---|
| Angle of Elevation | Angle between horizontal line and line of sight upwards. | |
| Opposite Side | Side opposite the angle in a right triangle. | Used in Sine and Tangent |
| Adjacent Side | Side next to the angle (not hypotenuse) in a right triangle. | Used in Cosine and Tangent |
| Tangent (tan) | Ratio of Opposite side to Adjacent side. | \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\) |
| \(\tan(30^\circ)\) | Specific value. | \(\frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\) |
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