Question

An airplane flies 200 km due west from city A to city B and then 325 km in the direction of 29.5° north of west from city B to city C.(a) In straight-line distance, how far is city C from city A?_____ km(b) Relative to city A, in what direction is city C?_____ ° north of west

Answer 1

We are given the following situation:

Part (a) we are asked to determine the distance from C to A. To do that we will use the cosine law, which is the following:

`a^2=b^2+c^2-2bc\cos A`

Where angle "A" is the angle opposite to side "a". In the given case, we have that:

`\begin{gathered} a=D_(CA) \\ b=325k\text{m } \\ c=200km \\ A=x \end{gathered}`

We can determine angle "x" using the fact that it is supplementary to the 29.5° angle, therefore, the add up to 180°:

`29.5+x=180`

Subtracting 29.5 from both sides we get:

`\begin{gathered} x=180-29.5 \\ x=150.5 \end{gathered}`

Now we substitute the values:

`D_(CA)^2=325^2+200^2-2(325)(200)\cos150.5`

We take the square root to both sides:

`D_(CA)=√(325^2+200^2-2(325)(200)\cos150.5)`

Solving the operations we get:

`D_(CA)=508.7`

Therefore, the distance from city C to city A is 508.7 km.

Part (b) We are asked to determine angle "y". To do that we will use the sine law:

`(\sin A)/(a)=(\sin B)/(b)`

Where "A" is the angle opposite to side "a" and "B" is the angle opposite to side "b".

Now, we substitute the values:

`(\sin150.5)/(508.7)=(\sin y)/(325)`

Now, we solve for "y". First, we multiply both sides by 325:

`(325)((\sin(150.5))/(508.7))=\sin y`

Now, we take the inverse function of the sine:

`\sin^(-1)((325)((\sin(150.5))/(508.7)))=y`

Solving the operations we get:

`18.34=y`

Therefore, the direction is 18.34° North of the west.