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:

$D_{CA}=\sqrt{325^2+200^2-2(325)(200)\cos150.5}$

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:

$\frac{\sin A}{a}=\frac{\sin B}{b}$

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

Now, we substitute the values:

$\frac{\sin150.5}{508.7}=\frac{\sin y}{325}$

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

$(325)(\frac{\sin(150.5)}{508.7})=\sin y$

Now, we take the inverse function of the sine:

$\sin^{-1}((325)(\frac{\sin(150.5)}{508.7}))=y$

Solving the operations we get:

$18.34=y$

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