test question
Asked by
| 10th Dec, 2009,
10:33: PM
In triangle ABC,
tan 45o = AB/ BC = 50/ BC
1 = 50/ BC
BC = 50
Let the height of the building be h.
Now, In triangle ADE
tan 30o = AE/ DE [AE = 50 - h, DE = 50m]
1/3 = (50 - h)/ 50
50/ 3 = 50 - h
h = 50 - 50/ 3
So, the height of building is 50 - 50/ 3
Horizontal distance of point of observation from top of the building = AD
cos 30o = DE/ AD
3 / 2 = 50 / AD
Therefore, AD = 100/3
Try on similar lines to get distance of top of cliff from the bottom of building.
Answered by
| 11th Dec, 2009,
11:01: AM
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