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Mathematics 17 Online
OpenStudy (anonymous):

Two towns, X and Y, are 25 km apart, with Y being due north of X. Town Z is due east of X and 60 km away. Draw a picture and find the distance and bearing of Y from Z??

OpenStudy (zzr0ck3r):

|dw:1403417147427:dw| \(?=\sqrt{25^2+60^2}\)

OpenStudy (anonymous):

Thank-you. How would i then find the bearing of Y from Z?

OpenStudy (anonymous):

Pythagorean Theorem, you have 2 sides and a right triangle. That is very important to learn for the future.

OpenStudy (anonymous):

I will be sure to remember it. What about the bearing?

OpenStudy (anonymous):

@Johnbc

OpenStudy (anonymous):

Bearing refers to the angle around the points we are looking at

OpenStudy (anonymous):

|dw:1403417909137:dw| For example

OpenStudy (anonymous):

Since you have a right angle triangle you can find the angles within the triangle and I believe those are the bearings

OpenStudy (anonymous):

Law of sines may be helpful \[\frac{ Sin A }{ a }=\frac{ SinB }{ b }=\frac{ SinC }{ c }\]

OpenStudy (anonymous):

I thought it was the other way around. |dw:1403418085937:dw|

OpenStudy (zzr0ck3r):

same thing

OpenStudy (anonymous):

Yes it can be both ways as long as you are consistent throughout

OpenStudy (anonymous):

oh ok

OpenStudy (zzr0ck3r):

\(\frac{a}{b}=\frac{c}{d}\iff\frac{b}{a}=\frac{d}{c}\)

OpenStudy (zzr0ck3r):

assume they are not 0

OpenStudy (anonymous):

ok

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