Question

From the top of a lighthouse , the angle of depression of two ships on the opposite sides of it are observed to be $\alpha$ & $\beta$ . If the height of the lighthouse be h metres & the line joining the ships passes through the foot of the lighthouse , show that the distance between the ships is $\frac {h(tan~ \alpha + tan~\beta)}{tan~\alpha~ tan~\beta}$ metres . ​

Answer 1

Question:

From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it is observed to be α &  β. If the height of the lighthouse be h meters & the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is  $\frac{h(tan \alpha +tan \beta)}{tan \alpha tan \beta}$ meters. ​

Answer:

Let PQ be the lighthouse whose given height is h meters.

See the attachment, here ∠PSQ=α and ∠PRQ= β (alternate interior angles)

Let's take PS=x and PR=y (for easy calculation)

it's a trigonometry question, so we will identify the right triangles in the figure which is ∠QPS and ∠QPR

In ∠QPS,

$cot \: \alpha = \frac{PS}{PQ} = \frac{x}{h} \\\\ \mapsto x= h cot\: \alpha$

And, In ∠PRQ,

$cot \: \beta = \frac{PR}{PQ} = \frac{y}{h} \\\\ \mapsto x= h cot\: \beta$

Now, the distance b/w two ships = RS = x+y

x+y= h cot α +h cot β

=  h (cot α + cot β)

$\mapsto h(\frac{1}{tan ~\alpha }+\frac{1}{tan ~\beta}) \\\\ \mapsto \frac{h(tan \alpha +tan \beta)}{tan \alpha tan \beta}$

Hence, Proved.

Answer 2

Answer:

really interesting xD

Step-by-step explanation:

I suck at math

( ･ั﹏･ั)

I've given up hope

or I would have buried my nose on maths book instead at this time