ht and distance(plz solve.....)
Asked by baidshryans
| 1st Oct, 2008,
10:19: PM
here let AB = 3a
then AP = a and PB = 3a
so length of the portion of the tree which is resting with the ground i.e. PC = 2a
now to calculate tanθ we have to find AC.
in right angle triangle APC , PC2 = AC2+AP2 ==> AC = 3a
so tanθ = AP/AC = a/3a = 1/
3 ==> θ = 30o
now we know AC we can find angle subtended by AB at C. let that be β
so tanβ = AB/AC = 3a/3a =
3
i.e. β = 60o
Answered by
| 2nd Oct, 2008,
12:23: AM
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