The angles of elevation of the top of a tower from two points distant a and b (a<b) from it's foot and the same straight line from it are 30° and 60°. show that the heights of the tower is ✓ab
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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
Step-by-step explanation:
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