Solve Physics Intro Homework Problem: Vector & Distance

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Two bicyclists are traveling to the same campground via different routes, with one cyclist first heading east and then north, while the other starts north and then heads directly toward the campground. The problem requires determining the distance of the second cyclist from the campground at the turning point and the direction they must travel relative to due east. The Pythagorean theorem is suggested for calculating distances, but initial attempts at applying it have been confusing. A sketch of the routes may help visualize the problem and clarify the calculations needed. Understanding the geometry involved is crucial for solving the assignment effectively.
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Hello, having a little trouble on an intro to physics homework problem.

Homework Statement



Two bicyclists, starting at the same place, are riding toward the same campground by two different routes. One cyclist rides 1050 m due east and then turns due north and travels another 1430 m before reaching the campground. The second cyclist starts out by heading due north for 2000 m and then turns and heads directly toward the campground.

Homework Equations


Pythagorean theorem?

The Attempt at a Solution


I tried using the theorem (used 2000 m as C^2 and 1050 m as A^2 since that is the distance that the first biker traveled), doesn't seem to work out.
I found out that the magnitude for the first biker is 1776? I believe that has to have some relevance to solving this problem.

Thanks for the help.
 
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What exactly is the question of the assignment?
 
Ok, ignore this.. I guessed (probably wrongly) what the question was!
 
oh stupid me..lol

the question is

At the turning point, how far is the second cyclist from the campground?

i guess if i can somehow figure the first one out i can figure the second part which is...

What direction (measured relative to due east) must the second cyclist head during the last part of the trip?
 
All you have to do is draw a 'trapezoidal' sketch and remember what a smart old greek said.
 

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