from an observation point the angles of depression of two boats in line with this point are found to 30° and 45°. find the distance between the...

Form a right triangle ABC  with the given information, such that

AB = height of the point of observation = `4000f`

BC=distance from the foot of the point to the first boat= `x` , `/_` ACB = `45^0`

Produce BC to D, such that `/_` ADB = `30^0`   , and DC =distance between two boats= `d m `

Now we evaluate these values by using trigonometric functions.

In `Delta` ABC

`tan 45^0 = (AB)/(BC) `

`1 = 4000/x `

`therefore x = 4000 `

In `Delta`ADB

`tan 30^0 = 4000/(x+d) `

`1/sqrt3 = 4000/(x+d) `

`x+d = 4000sqrt3 `

by substituting `x` value, we get 

`4000 + d = 4000sqrt3 `

`d = 4000(sqrt3 - 1) = 2928.4 f `

`therefore` the boats are `2928.4 f` apart.