From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the lighthouse be h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is h(tan α+tan β)tan α tan β metres. [3 MARKS]
Let AB be the lighthouse and C and D be the positions of the two ships. Then, AB = h metres.
Clearly, ∠ACB=α and ∠ADB=β
Let AC = x metres and
AD = y metres
From right ΔCAB, we have
ACAB=cot α⇒xh=cot α⇒x=h cot α ........... (i)
From right ΔDAB, we have
ADAB=cot β⇒yh=cot β⇒y=h cot β ........... (ii)
Adding the corresponding siddes of (i) and (ii), we get
x+y=h(cot α+cot β)=h(1tan α+1tan β)
⇒x+y=h(tan α+tan β)tan α tan β
Hence, the distance between the ships is h(tan α+tan β)tan α tan β m
Alternative Method,
In ΔDCB
tan α=hy
y=htan α [1 MARK]
In ΔDCA
tan β=hx
x=htan β [1 MARK]
The distance between the two ships =htan α+htan β
=h(tan β+tan αtan βtan α)m [1 MARK]