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Okay I have a math problem and I was wondering if anybody could help me with it. Here is the problem:

Josh takes a trip and travels 15 kilometers North, then he turn west and travels 16 kilometers, and then again he turns for 15 other kilometers North. I drew a diagram which has two triangles whose hypotenuses are kind of connected. I need to find the angle he travelled at. I'm not really sure how to do that.

2007-05-17 10:08:28 · 10 answers · asked by Anonymous in Science & Mathematics Mathematics

10 answers

28 degrees

Use the information given to show he travelled a total of 30 km north and 16 km west. The intersection of the north and west vectors is a right angle. Therefore, you know the "opposite" and "adjacent" legs of a right triangle that includes the angle you need. So the tangent of the angle is 16 km / 30 km = 8/15 and therefore the angle he travelled at is 28.07 degrees which rounds to 28 degrees in two significant digits.

2007-05-17 10:23:38 · answer #1 · answered by roynburton 5 · 0 0

You can ignore the fact that the northward distance was covered in separate trips. Josh ultimately traveled 30 km north and 16 km west. That's what you'll see if you connect the endpoints of his path. The tangent function should be able to give you the answer you're trying to find. Or, visually, take your diagram and draw additional horizontal and vertical lines to make the "kind of connected" hypotenuses be the hypotenuse of a single right triangle.

2007-05-17 10:13:27 · answer #2 · answered by DavidK93 7 · 0 0

If you are referring to the angle from where he started to where he ended, then you must make one triangle, with the one (north) side as 30 km (all the north distances) and the other (west) side as 16 km. Now the angle off of north would be found using tangent. So, using x as the angle,

tan x = 16/30
x = tan^(-1)(16/30) = 28.07 degrees

So the bearing would be N 28 degrees W. Hope this helps.

2007-05-17 10:17:21 · answer #3 · answered by Lee 3 · 0 0

Simply add the northward motion together, for a total of 30 km North, 16 km West and work out the trig from there. It should work out the same. Otherwise, use properties of parallel lines to determine the angle measurement. Without seeing a diagram, that's the best I can offer.

2007-05-17 10:14:33 · answer #4 · answered by Dark Knight 3 · 0 0

The easiest way is to treat it as one trianlge with a north-south leg of 30 km and an east-west leg of 16 km. Connect the starting point to the ending point. Then you can do a trig function on it.

ArcTan ( 15/16) will give the angle west of north he travelled.

43.152 degrees

2007-05-17 10:17:12 · answer #5 · answered by mr_moose_man 3 · 0 1

Think only about where he started and where he ends. If you chose his starting point to be (0,0), his overall trip took him 30 km north and 16 km west. So you have a right triangle with vertical leg of 30 and horizontal leg of 16. Use that triangle with your trig functions to find the angle of the total trip.

2007-05-17 10:13:38 · answer #6 · answered by apjok 3 · 0 0

Don't draw two triangles.

Draw one triangle with the total north distance (30) and the total west distance (16).

2007-05-17 10:13:24 · answer #7 · answered by Dr D 7 · 0 0

draw the diagram.OA=15 km vertically up(North)AB=16 km to the left and then BD=15 km vertically up again.let theta be the angle which OC makes with the vertical, then
tan(theta)=16/30=0.5333.
so theta=28.071degreesanwer

2007-05-17 10:25:06 · answer #8 · answered by Anonymous · 0 0

Hai Gowtham, Don't be afraid of trigonometry. the feeling of the fear is more dangerous. So, practise more problems on trigonometry.It will solve your problem. There is no substitution for Hard work. All the best.

2016-05-21 23:06:40 · answer #9 · answered by ? 4 · 0 0

when you create the two triagles the 16 km line is bisected into two equal parts...of which both equal 8

pythagorean
15 squared + 8 squared = 289
square root of 289= 17

now use sohcahtoa(sin oppo hyp cos adjacent hyp...etc)

sin x = 8/17
x=28.07 degrees

2007-05-17 10:20:54 · answer #10 · answered by Anonymous · 0 0

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