Asked by akheelfiroz | 16th Feb, 2010, 09:17: PM
Construct a right angled triangle, and mark one angle as 73 and other as 17.
Since cos 73 = m/n, the adjacent side would be m and hypotenuse would be n.
Hence the remaining side, (n2-m2).
cosec73 - cos17= 1/sin73 - cos17 = 1/[(n2-m2)/n] - (n2-m2)/n = (n2 - (n2-m2))/n(n2-m2) = m2))/n(n2-m2)
Since the angle of elevation from A is lower than B, it must be farther than location B.
AB = 50.
AP - BP = AB
AP - BP = 2BP
AP = 3 BP
BP = AB/2 = 50/2 = 25 ..............given.
AP = 75.
tanA = H/75 = ..........H - height of tower
tanB = H/25
tanB / tanA = 3
tan2A = 3 tanA
tan(A+A) = 3 tanA
2tanA/(1-tan2A) = 3 tanA
2 = 3 - 3tan2A
tanA = 1/3
A = 30, hence B = 60.
Height, H = APtan30 = 253 m
Answered by | 17th Feb, 2010, 07:02: AM
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