Question

From a window 60m high above the ground of a house in a street, the angle of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of opposite house is 60(1+√3) metres.

Answer 1

SOLUTION:

Let AP 60 m be the height of the window above the ground.(AP = QC)
CD = h m be the height of the house on the opposite side of the Street

GIVEN:
∠QPD = 60°(angle of elevation of the top of D of house CD)
∠QPC = 45°  (angle of depression of the foot C of the house CD)

QD = CD - CQ
QD = CD - AP        [CQ = AP]
QD =  (h - 60) m

In ∆PQC,
tan 45° = QC/PQ = P/B
1 = 60/PQ
PQ = 60 m

In ∆PQD ,
tan 60° = QD/PQ = P/B
√3 = (h-60)/60
60√3 = (h-60)
60√3 +60 = h
60(√3+1) = h

Hence, the height of the opposite house is 60(√3+1) m.

HOPE THIS WILL HELP YOU...

Answer 2

Hey....!! :))
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let AB be the height of the window and CD represent the height of the opp. house.Here, A represents the window and C & D are the top and foot of the opp. house. From A,draw a perpandicular AE to CD

So, CD=CE+ED

Also,AB=ED=60m

In triangle AEC, we have:-

tan60=CE/AE

or, root3=CE/AE

or, CE=AE x root3--------(1)

In tr. AED,we have:-

tan45=ED/AE

or, 1=60/AE

or,AE=60m-------(2)

Placing the value of AE from (2) to (1), we have :-

CE=60 x rt.3

Now, height of house=CD

But,CD=CE+ED

or, CD=60*rt.3+60

or, CD=60(rt.3 + 1)

Hence, we have showed that height of house is equal to

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I Hope it's help you....!!! :))