Angle of depression of boat from a 50m high Bridge is 30 degree then t...
Let the height of the bridge be “x” metres = 50 m
Let the distance between the bridge and the boat be “y” metres
Angle of depression is 30 degrees
=> angle of elevation form boat to the highest point of the bridge = 30 degrees
Tan 30 degrees = height of the bridge / distance between bridge and boat
=> (1 / √3) = x / y
=> 1 / √3 = 50 / y
=> y = 50√3
Distance of the boat from the bridge is 50√3 = 50 * 1.73 = 86.5 metres
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Angle of depression of boat from a 50m high Bridge is 30 degree then t...
The problem involves calculating the horizontal distance of a boat from a bridge, given the height of the bridge and the angle of depression.
Understanding the Scenario
- A bridge is 50 meters high.
- The angle of depression from the top of the bridge to the boat is 30 degrees.
Using Trigonometry
To solve for the horizontal distance (let’s call it "d") from the base of the bridge to the boat, we can use basic trigonometric concepts. The angle of depression is equal to the angle of elevation from the boat to the top of the bridge due to alternate interior angles.
Identifying the Right Triangle
- The height of the bridge (opposite side) = 50 meters.
- The angle of elevation = 30 degrees.
- The horizontal distance (adjacent side) = d.
Using the tangent function:
Formula
- The tangent of an angle in a right triangle is given by:
tan(θ) = opposite / adjacent
Substituting the values:
tan(30°) = 50 / d
Calculating d
- We know that tan(30°) = 1/√3 or approximately 0.577.
Thus, we set up the equation:
1/√3 = 50 / d
Solving for d
Cross-multiplying gives:
d = 50 * √3
- Using √3 ≈ 1.732:
d ≈ 50 * 1.732 ≈ 86.6 meters
Conclusion
Therefore, the horizontal distance of the boat from the bridge is approximately 86.6 meters.