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December 20, 2014

December 20, 2014

Posted by **Tammy** on Tuesday, March 15, 2011 at 2:38am.

b. The bearing they must travel along to get to base camp

- maths -
**MathMate**, Tuesday, March 15, 2011 at 10:09amI 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

Check all calculations please.

- maths -
**Tammy**, Tuesday, March 15, 2011 at 5:36pmThank you so much for all your help you my friend are a legend! Sorry about the mistakes :0

- maths :) -
**MathMate**, Tuesday, March 15, 2011 at 5:56pm:)

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