Problem 2
From the top of a cliff, 200 ft high, the angle of depression of the top and bottom of a tower are observed to be 30 and 60 degrees respectively. Find the height of the tower.
Solution
Let AB represent the tower and P the top of the cliff MP. If PX be the horizontal line through P, then angle XPA = 60 degrees. Let the height of the tower be
h ft.From A draw AL perpendicular to MP.
ML = AB =
h, therefore LP = (200-
h) ft
Again, angle PBM = XPB = 60 degrees; angle PAL = XPA = 30 degrees.
From rt. angle triangle PMB, BM/200 = cot(60)
Hence, BM = 200*cot(60)=200/(3)^1/2
From rt. angle triangle PLA, AL/LP = cot(30)
AL = LP*cot(30) = (200-
h)(3)^1/2
But, AL = BM
Therefore, (200-
h)(3)^1/2 = 200/(3)^1/2
or (200-
h)*3 = 200
3h = 400
h=400/3 ft
