Two lighthouses are located 25 miles from one another on a north-south line.

Question

Two lighthouses are located 25 miles from one another on a north-south line. If a person on a buoy is spotted 10° south of east from the northernmost lighthouse and 25° north of east from the southernmost lighthouse, how much closer is the closest lighthouse to the person on the buoy?

Answer 1

Hi Victoria!

When in doubt about something in geometry, I usually draw a picture. While I cannot do that in this explanation, you and I both can draw the triangle that is formed by the to lighthouses and the person on the buoy. This triangle has one vertical side of 25 miles, and two unknown sides; we'll call them A and B.

We know the angles of the triangle, from the directions from each of the lighthouses to the person.

Lighthouse N (for North) sees the person at 10 degrees south of east; that means that this is ninety minus ten equals eighty degrees east of south, and the angle between the N-S axis and this sightline is 80 degrees.

Lighthouse S (for South) sees the person at 25 degrees north of east. This is (90-25=65) degrees east of the N-S axis, and this angle is 65 degrees.

To find the angle between the two sightlines, add 80 and 65, and subtract this sum from 180. This gives

180-(80+65)=35 degrees, and we have all three angles of our triangle, but we still don't know sides A and B.

Use the Law of Sines to find A and B. This Law has that the sine of an angle, divided by the length of a triangle side across from it, is equal to a similar construction for each angle in that triangle. For your problem, this gives that:

sin(35)/25 = sin(65)/A = sin(80)/B, where all my numbers are in degrees.

Once you find A and B, find the difference.

Cheers,

Joseph F.