Two ships leave port at the same time and travel straight line distances, the first at 30 km/h and the second at 10 km/h. Two hours later they are 50 km apart. What is the angle between their courses?
3 answers
So you have three legs of a triangle. Perfect for the law of cosines. make a sketch, then use the law of cosines.
My triangle has sides 60 , 20 and 50 km
and I am finding the angle opposite the side of 50
by cosine law:
50^2 = 20^2 + 60^2 - 2(20)(60)cosØ
cosØ = .625
Ø = appr 51.3°
and I am finding the angle opposite the side of 50
by cosine law:
50^2 = 20^2 + 60^2 - 2(20)(60)cosØ
cosØ = .625
Ø = appr 51.3°
one goes 60 km
one goes 20 km
the third side is 50 km
divide them all by ten (similar triangle
then law of cosines
5^2 = 6^2 + 2^2 - 2*6*2 * cos A
25 = 36 + 4 - 24 cos A
-15 = -24 cos A
cos A = .625
A = 51.3
one goes 20 km
the third side is 50 km
divide them all by ten (similar triangle
then law of cosines
5^2 = 6^2 + 2^2 - 2*6*2 * cos A
25 = 36 + 4 - 24 cos A
-15 = -24 cos A
cos A = .625
A = 51.3