A ship sails 95 km on a bearing of 140 degree,then a further 102 km on a bearing of 260 degree and then returns directly to its starting point.find the length and bearing of the return journey.

Answer 1

I labeled my triangle ABC, with A as the starting point
AB = 95, and BC = 102 , and angle ABC = 60°
By the cosine law:
AC^2 = 98^2 + 102^2 - 2(95)(102)cos60°

find AC

Once you have AC, use the sine law to find angle BAC and
convert that into a bearing

Answer 2

why 98^2

Answer 3

I labeled my triangle ABC, with A as the starting point
AB = 95, and BC = 102 , and angle ABC = 60°
By the cosine law:
AC² = 95² + 102² - 2(95)(102)cos60
= 98.686
= 99 km