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Question: From the top of a lighthouse, the angle of the depression of two ships on the opposite side of it ar...

From the top of a lighthouse, the angle of the depression of two ships on the opposite side of it are observed to be α\alpha and β.\beta .If the height of the light house be hh meters and the line joining the ships passes through the foot of the light house, show that the distance between the ships is h(tanα+tanβ)tanαtanβmeters.\dfrac{{h(\tan \alpha + \tan \beta )}}{{\tan \alpha \tan \beta }}meters.

Explanation

Solution

To solve such problems. First draw a correct diagram taking ships as points and the lighthouse as a perpendicular line. Then convert the word problem into mathematical equations to solve the question. Use triangle properties of trigonometric ratios.

Complete step by step solution:
First, let us draw the diagram, explaining the question.

Let, A and B be two boats.
CD be the light house
XCA=α\angle XCA = \alpha and YCB=β\angle YCB = \beta be the angles of depression from the top of the light house to the boats A and B respectively.
We need to prove that the distance, AB between the ships is equal to h(tanα+tanβ)tanαtanβmeters.\dfrac{{h(\tan \alpha + \tan \beta )}}{{\tan \alpha \tan \beta }}meters.
Since XYAB,XY||AB,by using the property of alternate interior angles, we can say that
XCA=CAD=α\angle XCA = \angle CAD = \alpha and YCB=CBD=β\angle YCB = \angle CBD = \beta
Now, in ΔACD\Delta ACD
tanα=hAD\tan \alpha = \dfrac{h}{{AD}}
Rearranging it, we can write
AD=htanαAD = \dfrac{h}{{\tan \alpha }} . . . (1)
Similarly, In ΔBCD\Delta BCD
tanβ=hBD\tan \beta = \dfrac{h}{{BD}}
BD=htanβ\Rightarrow BD = \dfrac{h}{{\tan \beta }} . . . (2)
We know that the distance between the ships,
AB=AD+DBAB = AD + DB
Adding equation (1) and equation (2), we get
AB=AD+BD=htanα+htanβAB = AD + BD = \dfrac{h}{{\tan \alpha }} + \dfrac{h}{{\tan \beta }}
By taking hhcommon, we get
AB=h(1tanα+1tanβ)meterAB = h\left( {\dfrac{1}{{\tan \alpha }} + \dfrac{1}{{\tan \beta }}} \right)meter
By cross multiplying the terms, we can simplify it to
AB=h(tanβ+tanα)tanαtanβmeterAB = \dfrac{{h(\tan \beta + \tan \alpha )}}{{\tan \alpha \tan \beta }}meter
Hence, it is proved that the distance between the ships is h(tanα+tanβ)tanαtanβ\dfrac{{h(\tan \alpha + \tan \beta )}}{{\tan \alpha \tan \beta }} meters

Note: To speed up the calculation in the future, you can remember the fact that the angle of elevation and the angle of depression are always equal. Since, it was not given in the question that the ships are at the equal distance from the lighthouse, we did not assume AD=BD.AD = BD. You should keep this in mind to not assume something that is not given in the question.