An aircraft flies 400 km from a point O on a bearing of 025 degrees and then 700km on a bearing of 080 degrees to arrive at b. a) How far north of O is B? b) How Far East of O is B? c) Find the distance and bearing B from O
Answers
Answer:
Applying cosine law to OAB with A being the point where the aircraft changes its bearing to 080
OB = sqrt[400^2 + 700^2 - 2*400*700*cos(125)] = 985.5 and by law of sines
sin(AOB)/700 = sin(125)/OB => angle AOB = asin[sin(125)*700/985.5] =35.5803
The bearing of B from O is 25 + 35.5803 = 60.5803 degrees and OB = 985.5 km.
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Answer:
Applying cosine law to OAB with A being the point where the aircraft changes its bearing to 080
OB = sqrt[400^2 + 700^2 - 2*400*700*cos(125)] = 985.5 and by law of sines
sin(AOB)/700 = sin(125)/OB => angle AOB = asin[sin(125)*700/985.5] =35.5803
The bearing of B from O is 25 + 35.5803 = 60.5803 degrees and OB = 985.5 km.
Step-by-step explanation:
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