The distance of the top of the building from a is 163m and that from b is 115m .
Calculating the Perpendicular Distance of the Top From Ground
It is given that the angle of elevations of the building top from a and b are 25° and 37° respectively.
ab = 57m
Let the top of the building be point c and the distance between the building and the point b be x. Then, from the figure, we can deduce the following,
In Δacd, tan 25° = cd/ad
⇒ 0.4663 = cd /(x+57)
⇒ cd = 0.4663(x+57) ......................... (1)
In Δbcd, tan 37° = cd/x
⇒ 0.7536 = cd/x
⇒ x =cd / 0.7536 .......................... (2)
Solving for the Distance cd
Substitute the value of in equation (2) to equation (1) to get,
cd = 0.4663 ((cd/0.7536)+57)
cd = 0.4663(cd+42.9552)/0.7536
0.7536cd = 0.4663cd + 20.03
0.7536cd - 0.4663cd = 20.03
0.2906cd = 20.03
cd = 20.03/0.2906
Thus, the distance cd ≈ 68.93m
Finding the Distance cb and Distance ca
In Δacd, sin25° = cd/ca
⇒ 0.4226 = 68.93/ca
ca = 68.93/0.4226
ca = 163.10
∴ The distance ca ≈ 163m
Similarly, in Δbcd, sin37° = cd/cb
⇒ 0.6018 = 68.93/cb
cb = 68.93/0.6018
cb =114.54m
∴ The distance cb ≈ 115m
Therefore, the points a and b are at a distance of 163m and 115m respectively from the top of the building.
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