1.

From a window 20m high above the ground in a street, the angle of elevationand depression of the top and the foot of another house opposite side of thestreet are 60° and 45° respectively. Find the height of opposite house.( Take √ = 1.73)

Answer»

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.


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