question number 62...
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In the fig AB is the light house of height h (m)
In ∆ADC
tan β =AD/BC
tan β =h/y
h = y tan β or y =h/tanβ
In ∆ADB
tan α =AD/BD
tan α=h/x
h = x tan α or x = h/tan a
The distance between the two ships is BC = x + y
On adding eqn (1) & (2) we get,
x + y =h/tan a+h/tan b
h tanβ+h tan a/tan a tan b
h (tanβ+tan a)/tan a tan b metres
Proved.
In ∆ADC
tan β =AD/BC
tan β =h/y
h = y tan β or y =h/tanβ
In ∆ADB
tan α =AD/BD
tan α=h/x
h = x tan α or x = h/tan a
The distance between the two ships is BC = x + y
On adding eqn (1) & (2) we get,
x + y =h/tan a+h/tan b
h tanβ+h tan a/tan a tan b
h (tanβ+tan a)/tan a tan b metres
Proved.
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