The angle of elevation of the top of a building 30 m high from a point in the same plane is 60º, then the distance of the point from the foot of the building is
Question
The angle of elevation of the top of a building 30 m high from a point in the same plane is 60º, then the distance of the point from the foot of the building is
Solution
To solve this problem, we can use the tangent of the angle of elevation, which is the ratio of the opposite side (height of the building) to the adjacent side (distance from the building).
The problem gives us:
- The angle of elevation (60º)
- The height of the building (30m)
We need to find:
- The distance from the point to the foot of the building (let's call this d)
We know that tan(60º) = height/distance.
So, tan(60º) = 30/d.
We know that tan(60º) = √3.
So, √3 = 30/d.
Solving for d, we get d = 30/√3.
To rationalize the denominator, we multiply the numerator and the denominator by √3, so d = 30√3/3 = 10√3 m.
So, the distance of the point from the foot of the building is 10√3 meters.
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