Question

a car travels N28 degrees 20 minutes for 86 km and then N61 degrees 40 minutes W for 124 km what is its distance and bearing from the sticking point

Answer 1

someone answer pls

Answer 2

I made a sketch and marked all the angles
Let r be the distance between start and ending point.
the angle between the two trips is 90+28°20'+28°20' = 146° 40'
by the cosine law:
r^2 = 86^2 + 124^2 - 2(86)(124)cos146°40'
= 40591.28405
r = √...
= 201.477874.. km

angle of the triangle at start position:
sinØ/124 = sin146°40'/201.477874
sinØ = .3382...
Ø = 19.768° or 19°46'
so the angle using your notation
= 19°46' + 28°20' = 48°6
or N48°6'

using vectors:
|r| = (86cos28°20' , 86sin28°20') + (124cos61°40',124sin61°40')
= (75.6973.. , 40.8156) + (58.8504 , 109.14497)
= (134.54776.. , 149.9606...)
= √(134.54776..^2 + 149.9606..^2)
= 201.472787 , should have been exactly the same as above, probably typed in a decimal digit wrong
(I will let you find it, if you feel like it )

angle:
tan(angle) = 149.96.../134.54... = 1.11455..
angle = 48.1° = 48° 5'

again, that angle should be the same as above, I don't know where my arithmetic error is
(there is no error in the method)