Q)  From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be a and ẞ. If the height of the light house be h metres and the line joining the ships passes through the foot of the light house, show that the distance between the ship is \frac{h (\tan \alpha + \tan \beta)}{tan \alpha \\tan\beta} metres.

Ans: 

Step 1: Let’s draw a diagram for the given question:

From the top of a light house, the angles of depression of two ships

Let AB be the tower of height h, P and Q be the ships and angles of depression be α and β respectively.

We need to find the distance between the ships, D.

Here, ∠ P will be equal to α and ∠ Q will be equal to β (for being, alternate interior angles)

Step 2: In Δ  ABP, tan P = \frac{BA}{PA}

∴ tan α = \frac{h}{PA}

∴ PA = \frac{h}{\tan \alpha} ………….. (i)

Step 3: In Δ  ABQ, tan Q = \frac{BA}{AQ}

∴ tan β = \frac{h}{AQ}

∴ AQ = \frac{h}{\tan \beta} ………….. (ii)

Step 4: From the diagram, we can see that PQ = PA + AQ

By substituting values of PA & AQ from equations (i) and (ii), respectively, we get:

PQ = PA + AQ

∴ D = \frac{h}{\tan \alpha} + \frac{h}{\tan \beta}

∴ D = \frac{h \tan \alpha + h \tan \beta} {\tan \alpha \tan \beta}

∴ D = \frac{h (\tan \alpha + \tan \beta)} {\tan \alpha \tan \beta}

Hence Proved !

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