SOLUTION: From the top of a lighthouse 75m above sea level, the angles of depression of two buoys due north of the lighthouse are 60 degrees and 30 degrees respectively. Find the exact dista

Algebra ->  Trigonometry-basics -> SOLUTION: From the top of a lighthouse 75m above sea level, the angles of depression of two buoys due north of the lighthouse are 60 degrees and 30 degrees respectively. Find the exact dista      Log On


   

Question 1126068: From the top of a lighthouse 75m above sea level, the angles of depression of two buoys due north of the lighthouse are 60 degrees and 30 degrees respectively. Find the exact distance between the two buoys.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
in general, the angle of depression is equal to the angle of elevation.

therefore you have 2 triangles, both of which have a vertical side of 75 meters.

label the buoy with the 60 degree angle of depression C.
label the buoy with the 30 degree angle of depression D.
label the base of the lighthouse A.
label the top of the lighthouse B.

the length of AB is 75 meters.
the first buoy forms triangle ABC with the lighthouse.
the second buoy forms triangle ABD with the lighthouse.

angle ACB = 60 degrees.
anle ADB = 30 degrees.

tan(60) = AB / AC = 75 / AC.
tan(60) = sqrt(3), therefore:
sqrt(3) = 75 / AC.
solve for AC to get AC = 75 / sqrt(3) = (75 * sqrt(3)) / 3 = 25 * sqrt(3).

tan(30) = 75 / AD.
tan(30) = sqrt(3) / 3, therefore:
sqrt(3) / 3 = 75 / AD
solve for AD to get AD = 75 / (sqrt(3) / 3) = (75 * 3) / sqrt(3) = (75 * 3 * sqrt(3) / 3 = 75 * sqrt(3).

you have AC = 25 * sqrt(3) and AD = 75 * sqrt(3).

CD is equal to AD - AC = 75 * sqrt(3) - 25 * sqrt(3) = 50 * sqrt(3).

you have AC = 25 * sqrt(3) and CD = 50 * sqrt(3).

the distance between the buoys is equal to CD which is equal to 50 * sqrt(3).

my diagram is shown below:

$$$

the angle of depression is equal to the angle of elevation because they form
alternate interior angle to the parallel lines of BE and AD.