If the car has maintained a constant speed of 40 mi/h, how far is it from its starting position? (Round your answer to one decimal place.)

Question

A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If the car has maintained a constant speed of 40 mi/h, how far is it from its starting position? (Round your answer to one decimal place.)

Answer 1

The car starts out going east for 1 hour, which means it drove 40 miles east. Then, it turns northeast and travels for 30 minutes, which means it drove 1/2(40) = 20 miles northeast. Draw a non-right triangle with vertices at the car's starting position, the point where the car began driving northeast, and its current location. The vertex where the car began driving northeast is 180° - 45° = 135°, since a turn from east to northeast is 45°. The length of the side from starting point to northeast turn is 40 and the length of the side from northeast turn to current location is 20.

 

Since this isn't a right triangle, you can't use sine, cosine, and tangent like you usually do and you can't use the Pythagorean Theorem. If you've learned the Law of Cosines, you can use that to solve the problem, but based on the other questions you've posted to WyzAnt, I'll assume you haven't run into that yet. So, let's create a helpful right triangle to solve for our non-right triangle's third side (the side from the car's starting point to its current location).

 

The new triangle will have two vertices at the car's starting point and current location. The third vertex will be the point east of the car's starting point that's directly south of the car's current location. That vertex is the right angle.

 

Notice that this creates another, smaller right triangle as well, with vertices at the northeast turn, the car's current location, and the new point south of the car's current location. We'll find the legs of that smaller right triangle, whose non-right angles are 45° (because of the 45° turn from east to northeast). This triangle's hypotenuse is the car's northeast drive, so its length is 20. Use sine:

 

sine = opposite/hypotenuse

sin(45°) = x/20

0.707 = x/20

14.14 = x

 

The smaller right triangle's leg lengths are both 14.14 miles, since it's an isosceles triangle (both non-right angles are 45°).

 

Using this, we can tell that the new triangle's leg lengths are 14.14 and 14.14 + 40 = 54.14. Now, we can finally use the Pythagorean Theorem to solve for the new triangle's hypotenuse, which is the distance from the car's starting point to its current location. Let a = 14.14 and b = 54.14:

 

c² = 14.14² + 54.14²

c² = 199.94 + 2931.14

c² = 3131.08

c = 55.96

 

So, the car is approximately 56.0 miles from its starting position.

 

 

 

By the way, the Law of Cosines makes this problem much easier. You can use it on non-right triangles. It says that, in a triangle with sides a, b, and c where angle γ is across from side c...

 

c² = a² + b² - 2abcos(γ)

 

From the original triangle, plug in a = 40, b = 20, and γ = 135° and solve for c (the distance from the car's starting point to its current location):

 

c² = 40² + 20² - 2(40)(20)cos(135°)

c² = 1600 + 400 - 1600cos(135°)

c² = 2000 - 1600(-0.707)

c² = 2000 + 1131.37

c² = 3131.37

c = 55.96

 

Same answer as before, but much faster!

Answer 2

60 miles