SOLUTION: A girl starts from a point A and walks 285m to B on a bearing of 078�. she then walks due south u a point C which is 307m for A. what is the bearing of A from C, and what is |BC|

Question 1043857: A girl starts from a point A and walks 285m to B on a bearing of 078�. she then walks due south u a point C which is 307m for A. what is the bearing of A from C, and what is |BC|
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
|
|------------B
A------------|D
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|
|------------C
A to B is 285
That is the hypotenuse of the triangle ADB, which is a right triangle if AD has bearing 090 degrees.
Angle A is 12 degrees, since the bearing to B was 78 degrees.
From that we can get the value of BD. which is the sine of angle A.
sin A=BD/285
BD=285*sin A=59.25 m
CA has length 307
cos A is AD/AB; AD is AB cos 12=278.77 m
sin C=AD/AC=278.77/307=0.9080.
sin^(-1) (0.9080)=65.24 degrees. This bears west of north, so that the bearing from C to A is -65.24 degrees or 294.36 degrees. ANSWER
CD/AC is the cosine of angle C
cos 65.24=CD/AC
307 cos 65.24=CD
CD=128.58 m
Therefore BC=128.58+59.25=187.83 m ANSWER
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