two boat left a dock at the same time one travlled at 7km/h on a 39°. the other travelled at 5m/h on bearing of 82° haw far apart were the boats after 3 h round to the nearest tenths

Question

if you don't and put what formal you use

Answer 1

construct a triangle with one side = 7 km/hr x 3 = 21 km

another side is 5 m/hr x 3 = 15 meters = .015 kilometer Unless the k in km got accidentally left out, then that side = 15 km

the angle between these two side = 82-39 = 43 degrees

Use the Law of Cosines to find the 3rd side which is the distance the 2 boats are apart after 3 hours at the given speeds

c^2 = a^2 + b^2 -2abCosC

= 21^2 +15^2 -2(21)(15)Cos 43

= 441 +225 - 630Cos43

=about 666 + 460.75 = about 1126.75

c =square root of 1126.75 = about 33.567 km

assuming the 2nd boat's speed is 5 km/hr not 5m/hr

if it's really 5m/hr, work the problem the same way but keep the units the same, convert the km to m, or vice versa.

The Law of Cosines is a version of the Pythagorean Theorem, modified for triangles that are not right triangles