Question

An airplane flies 200 km due west from city A to city B and then 245 km in the direction of 36.0° 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

(a). 423.45 km

(b) 19.88 degrees north of west

Step-by-step explanation:

Let us sketch the displacement of the plane.

Part a.

From the above diagram, we can see that the straight line distance between A and C is the green line.

The angle ABC from geometry we find to be 180° - 36° = 144°.

Therefore, the law of cosines says

`CA^2=AB^2+BC^2-2(AB)(BC)\cos (144^o)`

Since AB = 200 and BC = 245, the above gives

`CA^2=200^2+245^2-2(200)(245)\cos (144^o)`

Which upon simplification gives

`CA^2=179308.6654\ldots`

Taking the square root and rounding to two decimal places gives

Hence, the straight line distance from A to C is 423.45 km.

Part b.

Finding the direction means we have to find the angle CAB in the above diagram. To do this, we again use the law of cosines.

Let us first rename CAB so that it is easy to write.

Let us call ∠CAB = x

Then the law of cosines gives

`BC^2=AB^2+CA^2-2(AB)(CA)\cos (x)^{}`

since BC = 245, AB = 200, and CA = 423.45, the above gives

`245^2=200^2+423.45^2-2(200)(423.45)\cos (x)`

then we solve for cos (x):

`\begin{gathered} 245^2-(200^2+423.45^2)=-2(200)(423.45)\cos (x) \\ \end{gathered}`
`\cos (x)=(245^2-(200^2+423.45^2))/(-2(200)(423.45))`

which we evaluate to get

`\cos (x)=0.94\ldots`

taking the inverse cosine gives

`x=\cos ^(-1)(0.94\ldots)_{}`
`x=19.88^o`

Or in other words,

`\boxed{\angle\text{CAB}=19.88^o\text{.}}`

part (c).

The answer to (a) and (b) are approximately correct because in each case we had to round our answer to appropriate decimal places. It is this rounding that makes us lose our accuracy in our answers.