Question

16. An airplane flies from city A to city B, a distance of 100 miles, and then turns through an angle of 20° and heads toward city C. If the distance from A to C is 300 miles, how far is it from city B to city C?

Answer 1

The student's question can be solved using the Law of Cosines to calculate the distance from city B to city C, with known distances and an angle. The provided step-by-step explanation outlines setting up the equation and inputting the known values to solve for the unknown distance.

Step-by-step explanation:

The student's question involves calculating the distance from city B to city C, given an airplane's flight path. We can use the Law of Cosines in this scenario because we have a triangle with one known side length (distance from A to C, which is 300 miles), an adjacent side (distance from A to B, which is 100 miles), and an included angle of 20° (the angle at which the airplane changes direction from B to C).

To find the distance from city B to city C, we can set up the equation from the Law of Cosines:

c² = a² + b² - 2ab • cos(θ)

where:

• c is the distance from B to C (what we are trying to find)
• a is the distance from A to B (100 miles)
• b is the distance from A to C (300 miles)
• θ is the angle the plane turns through when heading from B to C (20°)

Plugging the values into the equation, we get:

c² = 100² + 300² - 2 • 100 • 300 • cos(20°)

Then, we calculate c by taking the square root of the result.

Without providing a figure, we cannot compute the specific numerical answer. However, the method above outlines the correct approach to solving the problem.

Answer 2

Explanation

Step 1: We can find the measure of angle C using the law of sines.

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

Then, we have:

`\begin{gathered} a=? \\ b=300 \\ c=100 \\ A=? \\ B=160° \\ C=? \end{gathered}`
`\begin{gathered} (\sin B)/(b)=(\sin C)/(c) \\ (\sin(160°))/(300)=(\sin(C))/(100) \\ \text{ Multiply by 100 from both sides} \\ (\sin(160°))/(300)\cdot100=(\sin(C))/(100)\cdot100 \\ (\sin(160°))/(3)=\sin(C) \\ \text{ Apply the inverse function }\sin^(-1)\text{ from both sides} \\ \sin^(-1)((\sin(160°))/(3))=\sin^(-1)(\sin(C)) \\ 6.55°\approx C \\ \text{ The symbol }\approx\text{ is read 'approximately'.} \end{gathered}`

Step 2: We find the measure of angle A. For this, we can use the fact that the sum of the interior angles of a triangle is 180 degrees.

`\begin{gathered} A+B+C=180° \\ A+160°+6.55°\approx180° \\ A+166.55°\approx180° \\ \text{ Subtract 166.55\degree from both sides} \\ A+166.55°-166.55°\approx180°-166.55° \\ A\approx13.45° \end{gathered}`

Step 3: We find the measure of side a or that is the same, the distance between cities B and C. For this, we can use the law of sines again.

`\begin{gathered} \begin{equation*} (\sin A)/(a)=(\sin B)/(b) \end{equation*} \\ (\sin(13.45°))/(a)=(\sin(160°))/(300) \\ \text{ Apply cross product} \\ \sin(13.45°)\cdot300=a\cdot\sin(160°) \\ \text{ Divide by }\sin(160°)\text{ from both sides} \\ (\sin(13.45)\cdot300)/(\sin(160°))=(a\sin(160))/(\sin(160°)) \\ 204\approx a \end{gathered}`Answer

The distance between cities B and C is approximately 204 miles.