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 a 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 (2)
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 = 6 0 o
Using the cos rule, we can solve for the length x:
a 2 = b 2 + c 2 − 2 b ccos A x 2 = 3 5 2 + 2 5 2 − 2 ( 35 ) ( 25 ) cos 60 x 2 = 1850 − 875 x 2 = 975 x = 975 x = 31.2249...
Using the cos rule, we can solve for angle b:
a 2 = b 2 + c 2 − 2 b ccos A 3 5 2 = 31.22.. . 2 + 2 5 2 − 2 ( 31.22... ) ( 25 ) cos b 1225 = 975 + 625 − 1561.249... cos b − 375 = − 1561.249... cos b b = co s − 1 ( − 1561.249... − 375 ) 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)
The helicopter will need to fly approximately 58.2 km and follow a course of about 342.3 degrees to return to its home base. This distance and course were calculated using vector addition and the trigonometric principles. The calculations accounted for both legs of its journey to determine the direct route back.
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