Math Problem Statement

If Jane walks North for 3 miles, turns 45 degrees to the right, and then walks another 4 miles, how many miles will Jane be from her starting point? Give your answer as a decimal rounded to the nearest hundredth.

Solution

This problem can be solved using trigonometry and the Pythagorean theorem, since Jane's movements form a triangle.

  1. First leg of the journey: Jane walks 3 miles north.
  2. Second leg of the journey: After turning 45 degrees to the right, Jane walks 4 miles.

The two legs of her journey form two sides of a triangle, with the angle between them being 45°. We can now use the law of cosines to find the distance from Jane's starting point.

The law of cosines formula is: c2=a2+b22abcos(θ)c^2 = a^2 + b^2 - 2ab \cdot \cos(\theta) Where:

  • a=3a = 3 miles (first leg),
  • b=4b = 4 miles (second leg),
  • θ=45\theta = 45^\circ (the angle between the two legs).

Now, plug in the values: c2=32+422×3×4×cos(45)c^2 = 3^2 + 4^2 - 2 \times 3 \times 4 \times \cos(45^\circ) c2=9+1624×22c^2 = 9 + 16 - 24 \times \frac{\sqrt{2}}{2} c2=2524×0.7071c^2 = 25 - 24 \times 0.7071 c2=2516.97c^2 = 25 - 16.97 c2=8.03c^2 = 8.03 c=8.032.83c = \sqrt{8.03} \approx 2.83

So, Jane is approximately 2.83 miles from her starting point.

Would you like more details or have any questions?

Here are 5 questions you can explore:

  1. How would the result change if the second leg was 5 miles instead of 4 miles?
  2. What if Jane turned 60 degrees instead of 45 degrees after walking north?
  3. How would we solve the problem if Jane walked east instead of north at the start?
  4. Can you derive the law of cosines for any general triangle?
  5. How can this problem be solved using a vector approach?

Tip: For non-right triangles, always check if the law of cosines or sines is more suitable.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometry
Law of Cosines
Pythagorean Theorem

Formulas

c^2 = a^2 + b^2 - 2ab * cos(θ)

Theorems

Law of Cosines

Suitable Grade Level

Grades 10-12