[Guest]

You've traveled in a motorboat 8 miles away from the dock (the origin) at a 30^o angle above the shoreline (the x-axis). You then make a turn back toward the shore and travel exactly 5 miles so that your boat stops exactly on the shoreline.

Given the above information, you could be at exactly two different points on the shore (the x-axis) right now: (A,0) and (B,0). What is the distance (in miles) between these two points?

[HoustonHokie]

6 miles between the two points

[Guest]

Assuming that the shoreline is straight (which is a big assumption), the only way you can have two points that are 5 miles each from the shoreline is if you travel in an angled direction back to shore -- the same angle in either of the two possible directions. The straight line distance from the 8-miles-out point and the shoreline is 8*sin30º=4. Using the Pythagorean theorem, that gives a distance of 3 miles between this point and point A or point B. Thus, the distance between points A and B is 6 miles.

You've traveled in a motorboat 8 miles away from the dock (the origin) at a 30^o angle above the shoreline (the x-axis). You then make a turn back toward the shore and travel exactly 5 miles so that your boat stops exactly on the shoreline.

Given the above information, you could be at exactly two different points on the shore (the x-axis) right now: (A,0) and (B,0). What is the distance (in miles) between these two points?

[Guest]

boat travels 8 miles in one direction from shore and then returns returns 5 miles to shore, has travelled 2 sides of a right angled triangle! so distance from depart and arrival is X! 8²=5² + X² so X²= 64 - 25 X=6.245. Distance between the two possible positions is 12.49miles

Edited June 22, 2009 by bonanova
spoiler added

[Guest]

boat travels 8 miles in one direction from shore and then returns returns 5 miles to shore, has travelled 2 sides of a right angled triangle! so distance from depart and arrival is X! 8²=5² + X² so X²= 64 - 25 X=6.245. Distance between the two possible positions is 12.49miles

The only way this can be correct is if the boater did not initially travel in a 30° direction...

[Guest]

You've traveled in a motorboat 8 miles away from the dock (the origin) at a 30^o angle above the shoreline (the x-axis). You then make a turn back toward the shore and travel exactly 5 miles so that your boat stops exactly on the shoreline.

Given the above information, you could be at exactly two different points on the shore (the x-axis) right now: (A,0) and (B,0). What is the distance (in miles) between these two points?

[Guest]

Assuming that the shoreline is straight (which is a big assumption), the only way you can have two points that are 5 miles each from the shoreline is if you travel in an angled direction back to shore -- the same angle in either of the two possible directions. The straight line distance from the 8-miles-out point and the shoreline is 8*sin30º=4. Using the Pythagorean theorem, that gives a distance of 3 miles between this point and point A or point B. Thus, the distance between points A and B is 6 miles.

This is a poorly worded puzzle. With the information given your boat could be in 4 possible locations not 2. He said 30 deg above the shore. He did not specify a direction so 30 deg above the shore could be in either direction.

[Guest]

Here ya go...

6.0 miles between A and B.

[Guest]

This is a poorly worded puzzle. With the information given your boat could be in 4 possible locations not 2. He said 30 deg above the shore. He did not specify a direction so 30 deg above the shore could be in either direction.

No, it couldn't. the OP asked for the distance between the two end points, which is the same regardless of whether you travel for 8 miles with a five mile return starting bearing 60 or starting bearing 300 ("thirty degrees above the shore"). How could you possibly get to points C and D (if you start travel bearing 300) by travelling 8 miles bearing 60 with a five mile return trip?

On the other hand, the OP is obviously assuming no current

[Guest]

3.92 miles

Edited June 23, 2009 by rohit_bd