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.
Math - helper, Thursday, January 27, 2011 at 5:05pm
To understand this you really need to sketch this out.
Think of the x/y axis on a graph, with the y-axis as North and the x-axis as East. Both boats are traveling from the North (top of the y-axis), toward East (down toward the x-axis).
The first bearing, forms a 47 deg. angle with the y-axis and the second bearing, goes a little further forming a 79 deg. angle with the y-axis.
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