let C be the third vertex of the horizontal triangle, directly below the building
for the vertical triangles, height = AC(tan(45)) = BC(tan(30)) = AC, since tan(45) = 1
now use law of cosines on the horizontal triangle to obtain
250^2 = AC^2 + (AC/tan(30))^2 – 2(AC)(AC/tan(30))cos(135)
and solve the quadratic equation to obtain AC = height = 98.4413
From A, due east of a building, the angle of elevation of the top is 45 degree. From B, southwest of the building , the angle of elevation is 30 degree. If AB=250m, find the heigt of the building.
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