Asked by Precious
A girl walks at a distance of 2km and on a bearing of 107° at a point A to B,she changes direction from B to C,on a bearing of 017° and at a distance of 3km,what is the bearing from C to A?
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Answered by
Anonymous
A girl walks at a distance of 2km and on a HEADING of 107° at a point A to B,she changes direction from B to C,on a HEADING of 017° and at a distance of 3km,what is the bearing (YES, finally, ,, math books do not do navigation) from C to A?
anyway
107 = 17 deg below +x axis
17 = 17 degrees clockwise from +y axis (NOTE PERPENDICULAR to AB)
So ABC is a right triangle with 90 deg angle at B
so what is angle BAC ???
Tan BAC = 3/2 = 1.5
So BAC = 56.3 deg
so
AC = 56.3 - 17 = 39.3 deg above +x axis
so
Line AC is 90-39.3 = 50.6 deg clockwise from north
so
CA = AC + 180 = 230.6 deg clockwise from north
anyway
107 = 17 deg below +x axis
17 = 17 degrees clockwise from +y axis (NOTE PERPENDICULAR to AB)
So ABC is a right triangle with 90 deg angle at B
so what is angle BAC ???
Tan BAC = 3/2 = 1.5
So BAC = 56.3 deg
so
AC = 56.3 - 17 = 39.3 deg above +x axis
so
Line AC is 90-39.3 = 50.6 deg clockwise from north
so
CA = AC + 180 = 230.6 deg clockwise from north
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