a hiker travels 3 km due east then 2km on a bearing of 110° how far south is the hiker from the starting point

Answer 1

due east is 90º

110º is 20º south of east

distance south of the start is ... 2 km * sin(20º)

Answer 2

Draw a horizontal line.

Mark start point as A.

Mark the point after 3 km as B.

At that point B, draw an angle θ = 110 ° with respect to the horizontal.

Mark the point after 2 km as C.

∠ θ = 110° = ∠ ABC

∠ is the mark for angle

Mark the length AB as a.

a = 3 km

Mark the length BC as b.

b = 2 km

Mark the length AC as c.

Now law of cosines:

c² = a² + b² - 2 ∙ a ∙ b ∙ cos θ

c² = a² + b² - 2 ∙ a ∙ b ∙ cos 110°

c² = 3² + 2² - 2 ∙ 3 ∙ 2 ∙ cos 110°

c² = 9 + 4 - 12 ∙ cos 110°

c² = 13 - 12 ∙ ( - 0.342020143 )

c² = 13 + 4.104241716‬

c² = 17.104241716‬

c = √17.104241716‬

c = 4.1357274712 km

Answer 3

"how far south is the hiker from the starting point"
means that you only want the y co-ordinate of the resulting vector

(3cos0 , 3sin0) + (2cos(-20), 2sin(-20))
= (3,0) + (1.8794, -.684)
= (4.8794, -.684)

So the southward position from his starting point is .684 km

If we want the distance from his starting points we get
√(4.8794^2 + (-.684)^2) = 4.927 km

If you change the angle in Bosnian's solution to 160°, we get the same distance answer.