SOLUTION: Please help with this question,I'm really stock.I need this tomorrow.Thank you..
A cross country race course on flat ground has runners,Jedd and Rodney, heading from point A for 3
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Length-and-distance
-> SOLUTION: Please help with this question,I'm really stock.I need this tomorrow.Thank you..
A cross country race course on flat ground has runners,Jedd and Rodney, heading from point A for 3
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Question 438239: Please help with this question,I'm really stock.I need this tomorrow.Thank you..
A cross country race course on flat ground has runners,Jedd and Rodney, heading from point A for 3.5km on a bearing of 047degrees temperature to point B. From B they turn and run for 2.8km on a bearing of 342degrees temperature to point C. They then run back to point A.
(a.) Determine the distance from C back to A.
(b.) At what bearing does the course follow from C back to A? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A cross country race course on flat ground has runners,Jedd and Rodney, heading from point A for 3.5km on a bearing of 047degrees temperature to point B. From B they turn and run for 2.8km on a bearing of 342degrees temperature to point C. They then run back to point A.
(a.) Determine the distance from C back to A.
(b.) At what bearing does the course follow from C back to A?
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These degrees have nothing to do with temperature ??? They're compass headings. No wonder you're "stock."
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0 degrees is North. 047 degs is 47 degrees to the East of North. If you make a sketch, you'll see that angle CBA = 115 degrees (not a heading)
Now you have a triangle with 2 sides and the included angle.
Use the Cosine Law to find side AC, side b.
b =~ 5.32666 km
= distance from C to A
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Find angle C using the Law of Sines:
b/sin(115) = 3.5/sin(C)
sin(C) = 3.5*sin(115)/b =~ 0.5955
Angle C = 36.55 degs (not a heading)
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The heading from A to C = 047 - 36.55 =~ 010
From C to A is the reciprocal,
Heading = 190 degrees