The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the sqme straight line with it are complementary. Prove that the he8ght of the tower is √ab metres
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From the Figure,
In ΔACD,
TanΘ = CD/AD
= CD/a ................(i)
In ΔBCD,
Tan(90-Θ) = CD/BD
= CD/b ................(ii)
Multiplying (i) and (ii)
TanΘ × Tan(90-Θ) = CD/a × CD/b
TanΘ × CotΘ = CD²/ab
(because Tan(90-Θ) = CotΘ)
1 = CD²/ab
(because TanΘ × CotΘ = 1)
CD² = ab
CD = √ab metres
Hence the Height = √ab metres
Hence Proved.
_______________________________________________________
☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
In ΔACD,
TanΘ = CD/AD
= CD/a ................(i)
In ΔBCD,
Tan(90-Θ) = CD/BD
= CD/b ................(ii)
Multiplying (i) and (ii)
TanΘ × Tan(90-Θ) = CD/a × CD/b
TanΘ × CotΘ = CD²/ab
(because Tan(90-Θ) = CotΘ)
1 = CD²/ab
(because TanΘ × CotΘ = 1)
CD² = ab
CD = √ab metres
Hence the Height = √ab metres
Hence Proved.
_______________________________________________________
☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
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