The angle of the elevation of the top of a tower from two points on the ground at a distance is 'a' metre and 'b' metres from the base of the tower and in the same straight line with it are complementary. Then the square of the height of the tower is :-
Answers
Answer:
Step-by-step explanation:
Given,
the angle of elevation of the top of the tower from two points P & Q is at a distance of a & b.
Also given, to prove that the tower
height=
ab
(∵ complementary angle =(90
o
−θ))
From ΔABP
tanθ=
BP
AB
=
a
AB
……..(1)
From ΔABQ
tan(90−θ)=
BQ
AB
(∵tan(90−θ)=cotθ)
(cotθ=
tanθ
1
)
We get,
cotθ=
AB
BQ
=
AB
b
……..(2)
by equation (1) & (2) we get,
a
AB
=
AB
b
⇒AB
2
=ab⇔AB=
ab
∴AB=height=
ab
Hence proved.
Answer:
Given,
the angle of elevation of the top of the tower from two points P & Q is at a distance of a & b.
Also given, to prove that the tower
height=
ab
(∵ complementary angle =(90
o
−θ))
From ΔABP
tanθ=
BP
AB
=
a
AB
……..(1)
From ΔABQ
tan(90−θ)=
BQ
AB
(∵tan(90−θ)=cotθ)
(cotθ=
tanθ
1
)
We get,
cotθ=
AB
BQ
=
AB
b
……..(2)
by equation (1) & (2) we get,
a
AB
=
AB
b
⇒AB
2
=ab⇔AB=
ab
∴AB=height=
ab
Hence proved.
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