# Answer to Question #15824 in Mechanics | Relativity for Cecilia

Question #15824

Part 1: An airplane ď¬‚ies 221 km due west from city A to city B and then 387 km in the direction of 25.5 degrees north of west from city B to city C. What is the distance between city A and city C?

Answer in units of km

Part 2: Relative to city A, in what direction is city C? Answer with respect to due east, with the

counter-clockwise direction positive, within the limits of −180 degrees to +180 degrees.

Answer in units of degrees

Answer in units of km

Part 2: Relative to city A, in what direction is city C? Answer with respect to due east, with the

counter-clockwise direction positive, within the limits of −180 degrees to +180 degrees.

Answer in units of degrees

Expert's answer

Part 1:

From the theorem of cosines we get for triangle ABC: AB^2+BC^2-ABBCcos(B)=AC.

After putting numbers we get answer. 479 km.

Part 2:

If we take BC as hypotenuse of triangle ABH with angle H=90, then BC*sin(180-115)=CH

CH will be leg of triangle ACH, where we don't know angle A, but when we use

definition of tangents, we will get AB+BH=Ab+BC*Cos(180-115) and AC known from the

task.

Arctg(AC/(AB+BH) gives us 42.7 degrees

If we calculate due to east, it will be 137.2 degrees.

From the theorem of cosines we get for triangle ABC: AB^2+BC^2-ABBCcos(B)=AC.

After putting numbers we get answer. 479 km.

Part 2:

If we take BC as hypotenuse of triangle ABH with angle H=90, then BC*sin(180-115)=CH

CH will be leg of triangle ACH, where we don't know angle A, but when we use

definition of tangents, we will get AB+BH=Ab+BC*Cos(180-115) and AC known from the

task.

Arctg(AC/(AB+BH) gives us 42.7 degrees

If we calculate due to east, it will be 137.2 degrees.

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