ht and distance(plz solve.....)

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 = 30^o

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 = 60^o