Trig Math Functions

Question

The crows nest of the yacht A is 50 meters above the water. The angle of depression from the nest to a buoy due west is 40 degrees, and to a second buoy of 34 degrees which is located S70degreesW of the yacht. How far apart are the buoys?

Answer 1

Answer 2

You're working with 3 dimensions so I drew it with 2views.

First I drew a side using the 50m, 40° and 34°. Now there's

2 radiuses.

For the top view I set the center; I drew the circles with the

2 rad. Here I made the triangle which is the top of the water.

Now that I drew the rads; I need the distances. Looking at

the side view; side "a1"=50; "β1"=90-40=50°; α1=40°; γ1=90°. Want side "b".

a/sinα=b/sinβ=c (sine of a rt >=1).

50/sin40=70.78m=c1 (1st buoy).

Now we want c2.

Side view: a2=50; α2=34°; β2=56°; γ2=90°.

50/sin34°=89.41=c2=b/sin56°; b=74.128.

Back to the top view we want side "a". The formula is b^2+c^2-2bc(cosα)=a^2.

70.78^2+74.13^2-(2*70.78*74.13).9396=25.378m (2nd buoy)