# maths

posted by Tammy

A party of bush walkers have an average speed of 50km/minute and walk for 4 hours on a bearing of 231°T. they then walk for 6(1)/(3) hours on a bearing of 117°T. Find a.How far are they from bas e camp?
b. The bearing they must travel along to get to base camp

1. MathMate

I assume the speed of the adventurers are 50 m/min. since 50km/min is almost the speed of a supersonic jet.

Also, not clear about the 6(1)/(3) hours. I take it at face value of 2 hours.

The bearing of 231°T will be assumed to be measured from true north, clockwise, equivalent to "azimuth" used in US and French speaking countries.

The solution here lies in the resolution of all distances travelled to directions east (x-axis) and north (y-axis). The components can then be summed together to get the final destination and direction, if required.

There is a little problem combining bearings and trigonometry, the latter measures angles from zero in the east direction counter clockwise.

So to calculate sines and cosines using trigonometry, and with x≡East and Y≡North, we subtract the bearing from 90°.

For example, a bearing of 117° will be calculated as (90-117)=-27° in trigonometry.

Now we're ready to sum the distances:

The distances are:
1. 4*60*50=12000m
2. 2*60*50=6000m

Bearing Angle
231°T -> 90-231=-141°
117°T -> 90-117=-27°

leg Distance(D) Angle(A) Dcos(A) Dsin(A)
1 12000 -141° -9325.75 -7551.84
2 6000 -27° 5346.04 -2723.94
Sum (x2,y2)=(-3979.71,-10275.78)
Distance = sqrt(x2²+y2²)
angle
= arctan(-10275.78/-3979.71)
Note that since both numbers are negative, angle is in 3rd quadrant
= 248.83°(trig.)
convert to bearing, subtract from 90 and add 360 if result is negative
= (90-248.83)+360
= 201.17°T

2. Tammy

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