A road starts at a college and goes due north for 2000m, it then goes 2000m on a bearing 040 degree and ends at a market. How far is the market from the college

Question

A picturesque scene depicting an educational institution situated at the southern end of a straight road. This road should stretch as far as the eye can see towards the north, encompassing a distance of 2000m. This road then abruptly takes a turn towards the right, maintaining the same stretch of 2000m, and finally leading us to a bustling market place. The angle between the two roads is 40 degrees. The setting is a clear day with a bright sun overhead, budding trees along the streets and a slew of people bustling back and forth on the road and in the market.

A road starts at a college and goes due north for 2000m, it then goes 2000m on a bearing 040 degree and ends at a market. How far is the market from the college

Answer 1

arghhhh!

my angle should have been 140°, but I am sure you had that
figured out.

Answer 2

All angles are measured CCW from +x-axis.

d = 2km[90o] + 2km[140o].
X = 2*Cos90 + 2*Cos140 = 0 - 1.53 km. = -1.53 km.
Y = 2*sin90 + 2*sin140 = 2 + 1.29 = 3.29 km.

d = -1.53 + 3.29i = 3.63km[-65o] = 3.63km[65o] N. of W. = 3.63km[115o].
3.63 km = 3630 m.

Answer 3

I will assume you made a sketch

by the cosine law:
x^2 = 2^2 + 2^2 - 2(2)(2)cos100°

solve for x

multiply the answer by 1000, since I scaled it back
careful with that last term, remember cos100° is negative.

Answer 4

Well, if the road went due north for 2000 meters and then took a 040-degree turn, it's safe to say that the market is pretty far from the college. In fact, it's so far that the college students might need a nap along the way! Just kidding!

To find the distance between the college and the market, we can use some trigonometry. Let's break it down:

The road goes due north for 2000 meters, which we can call the vertical distance. Then, it takes a 040-degree turn, which gives us the angle between the road and the north direction.

Using trigonometry, we can find the horizontal distance by multiplying the vertical distance (2000m) by the cosine of the angle (040 degrees).

So, the market is approximately 1533.6 meters away from the college (if we round to one decimal place). But remember, this is just a rough estimate for a good laugh!

Answer 5

To determine the distance from the college to the market, we can break down the road into two segments: the northward segment and the segment on a bearing of 040 degrees.

1. Northward Segment: The road goes due north for 2000 meters. Since it's a straight line, the distance is simply 2000 meters in this direction.

2. Segment on a Bearing: The road then goes 2000 meters on a bearing of 040 degrees. To find the distance traveled in this direction, we need to calculate the horizontal component of the displacement.

The horizontal component is given by the formula: horizontal distance = distance * cos(angle)

Applying this formula, we get:
horizontal distance = 2000m * cos(40 degrees)

Using a calculator or trigonometric table, we find that the cosine of 40 degrees is approximately 0.766.

So, the horizontal distance is:
horizontal distance = 2000m * 0.766 ≈ 1532 meters

Now, we can find the total distance from the college to the market by summing up the distances:
Total distance = Northward Segment + Segment on a Bearing
Total distance = 2000 meters + 1532 meters
Total distance ≈ 3532 meters

Therefore, the market is approximately 3532 meters away from the college.