Question

A rescue helicopter flew from its home base for 35 kilometers on a course of 330 degrees to pick up an accident victim. It then flew 25 kilometers on the course of 90 degrees to the hospital. what distance and on what course will the helicopter fly to return directly to its home base?

Answer 1

I've drawn a triangle to represent the journey (see attached.

The angle from Home to Victim to Hospital is:

a = $90 - (360 - 330) = 90- 30 = 60^o$

Using the cos rule, we can solve for the length x:

$a^2 = b^2 +c^2 - 2bccosA \\x^2 = 35^2+25^2-2(35)(25)cos60 \\x^2 = 1850 - 875 \\x^2 = 975 \\x = \sqrt{975} \\x = 31.2249...$

Using the cos rule, we can solve for angle b:

$a^2 = b^2 +c^2 - 2bccosA \\35^2 = 31.22...^2+25^2-2(31.22...)(25)cosb \\1225 = 975 + 625 - 1561.249...cosb \\-375 = -1561.249...cosb \\b = cos^{-1}( \frac{-375}{-1561.249...}) \\b = 76.1...$

So the bearing the helicopter has to travel is:

270 - 76.1 = 193.9 degrees (1 dp)

And the distance it travels is 31.2 km (1 dp)