Math
posted by helpisneeded .
Two Boats left the harbour at the same time. One travelled at 10km/h on a bearing of N47°E. The other travelled at 8km/h on a bearing of N79°E. How far apart were the boats after 78 mins? Round the distance to the nearest tenth of a kilometer. Use trigonometry.

To understand this you really need to sketch this out.
Think of the x/y axis on a graph, with the yaxis as North and the xaxis as East. Both boats are traveling from the North (top of the yaxis), toward East (down toward the xaxis).
The first bearing, forms a 47 deg. angle with the yaxis and the second bearing, goes a little further forming a 79 deg. angle with the yaxis.
The triangle formed, (I labeled ABC), with A at the origin, side c (between angle A and angle B), side b (between angle A and angle C) and side a (between angle B and angle C).
You need to find side a, the distance between the boats.
1 km/h / 60 min = 0.01667 km/min
First boat 10 km/h * 0.01667 = 0.1667 km/min * 78 min = 13 km
Second boat 8 km/h * 0.01667 = 0.1333 km/min * 78 min = 10.4 km
Angle A = 79 deg  47 deg = 32 deg.
side c = 13 km (first boat's bearing)
side b = 10.4 km (second boat's bearing)
To find side a,
a^2 = b^2 + c^2  2bc cos A
a^2 = 10.4^2 + 13^2  2(10.4)(13)cos 32d
a^2 = 108.16 + 169  270.4 (.8)
a^2 = 277.16  216.32
a^2 = 60.8 km