Explanation
In ΔDPC
tanp=CDCP
tan300
1√3=CDCP
CP=CP√3 (1)
IN ΔAPB
tanp=ABPB
tan600=ABBP
√3BP=AB
CD=√3BP (2)
FROM (1) & (2)
CP√3=√3BP
CP=3BP
NOW ,BC=BP+CP
80=BP+3BP
BP=20m
CP=60m
CD=√3BP
=20√3
Let the distance between point of observation and foot of tower BC=20m
Height of tower AB=H
Angle of elevation of top of tower θ=600
Now from figure ΔABC
We know that, ΔABC is a right angle triangle.
tanθ=ABBC
tan600=h20
⇒h20=√3
⇒h=20√3m.
Step - 1: Find tan α
H = height of the tower
α = from bottom of a pole the angle of elevation the top of a tower
β = subtends angle
Let AO = BM = a and AB = OM = h.
∴PM = H - h
In ΔAOP
tanα = POAO
⇒AO = Hcotα
⇒BM = Hcotα......(i)
Step -2: Find the height of the tower.
In ΔPBM,tan(90−α+β)=BMPM
⇒BM = PM tan(90−α+β)
=(H - h)cot(α−β)…..(ii) by using [tan(90−θ) = cotθ]
From (i) and (ii),
Hcotα=(H - h)cot(α−β)
⇒hcot(α−β) = H[cot(α−β)−cotα]
⇒H = hcot(α−β)cot(α−β)−cotα.....[Proved]
Hence the correct answer is option B
A person , standing on the bank of a river observer that the angle subtended by tree on the opposite bank is, when he retreats 40 meters from the bank, he finds the angle to be . the height of the tree and the breadth of the river are :
Please disable the adBlock and continue. Thank you.