Math Problem Statement

The bearing from A to C is S 51​° E. The bearing from A to B is N 88​° E. The bearing from B to C is S 39​° W. A plane flying at 250 mph takes 3.6 hr to go from A to B. Find the distance from A to C.

Solution

To solve this problem, we can use the law of sines and some trigonometric identities based on the given bearings. Here's a step-by-step approach:

Step 1: Analyze the Bearings

  • The bearing from A to C is S51ES 51^\circ E, which means the angle measured clockwise from the south is 5151^\circ.
  • The bearing from A to B is N88EN 88^\circ E, which means the angle measured clockwise from the north is 8888^\circ.
  • The bearing from B to C is S39WS 39^\circ W, which means the angle measured clockwise from the south is 3939^\circ.

Step 2: Calculate the Angle BAC\angle BAC

To find the angle BAC\angle BAC, we consider the difference in direction between the bearings from A to C and from A to B:

  • Bearing N88EN 88^\circ E from A to B corresponds to an angle of 8888^\circ clockwise from the north.
  • Bearing S51ES 51^\circ E from A to C corresponds to an angle of 18051=129180^\circ - 51^\circ = 129^\circ clockwise from the north.

So, the angle BAC=12988=41\angle BAC = 129^\circ - 88^\circ = 41^\circ.

Step 3: Calculate the Angle ABC\angle ABC

Now, calculate the angle ABC\angle ABC between the directions from A to B and B to C:

  • Bearing S39WS 39^\circ W from B to C corresponds to an angle of 180+39=219180^\circ + 39^\circ = 219^\circ clockwise from the north.
  • Bearing N88EN 88^\circ E from A to B corresponds to an angle of 8888^\circ clockwise from the north.

So, the angle ABC=21988=131\angle ABC = 219^\circ - 88^\circ = 131^\circ.

Step 4: Calculate the Remaining Angle BCA\angle BCA

Since the sum of the angles in a triangle is 180180^\circ, the remaining angle BCA\angle BCA is: BCA=180BACABC=18041131=8\angle BCA = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 41^\circ - 131^\circ = 8^\circ

Step 5: Calculate the Distance from A to B

The plane travels from A to B at 250 mph for 3.6 hours. Therefore, the distance from A to B is: AB=250 mph×3.6 hours=900 milesAB = 250 \text{ mph} \times 3.6 \text{ hours} = 900 \text{ miles}

Step 6: Use the Law of Sines to Find AC

Using the Law of Sines: ACsin(ABC)=ABsin(BCA)\frac{AC}{\sin(\angle ABC)} = \frac{AB}{\sin(\angle BCA)} Substitute the known values: ACsin(131)=900sin(8)\frac{AC}{\sin(131^\circ)} = \frac{900}{\sin(8^\circ)} AC=900×sin(131)sin(8)AC = \frac{900 \times \sin(131^\circ)}{\sin(8^\circ)} Now, calculate the values: AC900×0.75470.1392679.230.13924879.25 milesAC \approx \frac{900 \times 0.7547}{0.1392} \approx \frac{679.23}{0.1392} \approx 4879.25 \text{ miles}

Thus, the distance from A to C is approximately 4879.25 miles.

Would you like more details, or do you have any other questions?


Here are 5 related questions:

  1. How would the calculation change if the plane's speed varied during the flight?
  2. Can the law of sines be used to find the other distances in the triangle? How?
  3. How would you determine the distance if the bearings were given in a different format?
  4. What is the significance of bearings in navigation, and how are they typically measured?
  5. What trigonometric laws could be used if the problem involved an obtuse triangle?

Tip: Always double-check the bearings and angles when working with navigation problems, as small errors can lead to significant inaccuracies.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Law of Sines
Bearings

Formulas

Law of Sines: \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \)

Theorems

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Suitable Grade Level

Advanced High School