Think of the explorer’s position at each stage in the expedition as a distance north (N) from the starting point, and a distance east (E). Because they are off-course to the east, they have to travel further north and come back to the west (W).

After 3 days, the distances travelled in km are:

N: 120cos(7) …(1)

E: 120sin(7) ….(2)

During the next 3 days, the additional distances are:

N: 120cos(10) ….(3)

E: 120sin(10) ….(4)

The total distances travelled are:

from (1) & (3): N: 120(cos(7) + cos(10))

from (2) & (4): E: 120(sin(7) + sin(10))

The distances remaining to travel are:

N: 400 – 120(cos(7) + cos(10)) = 162.72km.

W: 120(sin(7) + sin(10)) = 35.46km.

Combining these distances with Pythagoras’ Theorem (as they are perpendicular to each other), the total straight line distance is:

sqrt(162.72^2 + 35.46^2)

= 166.54km.

The direction is N x W where:

tan(x) = 35.46 / 162.72

x = 12.29deg.

They must travel a further 166.54km in a direction

N 12.29deg W.

2.

In 2hr, the trio has walked 3km/hr * 2hr. = 6km.

If x is the distance E from the first team member’s course which is to the N, then:

x / 6 = sin(65)

x = 6sin(65)

x = 5.44km.