# Thread: bearings and cartesian plane

1. ## bearings and cartesian plane

hello some questions for bearings I have to draw these manually.

a man walks 3.6 kilometres on bearing 235 degrees from A to B.
draw a diagram and find how far B is west of A and how far B is south of A?
How to find the angles in the triangle?

a ship sails on bearing 324 degrees from port p to port q a distance of 160 nautical miles.It then sails east until it reaches port r , due north of p.
What is the bearing of port p from port q?
what is the shortest distance from port r to port p?

a ship sails from x to y on a bearing of 125 degrees.if x to y is 234 nm how far is y south of x?

cartesian plane
draw a diagram of the cartesian plane to show point p (1,3) and point q (-3,-2)
I know how to find gradient, midpoint and length but
draw a line 5x-4y-4=0
and explain why this is parallel to pq.
in the diagram calculate the angle made by line pq and the positive x axis.

a graph shows
time 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0
temp C 54, 48, 42, 35, 30, 27 , 22

draw a scatter diagram and show line of best fit.
find the vertical axis intercept
write down the equation of the line describing the linear relationship between the temperature of the liquid and time of cooling.
use the equation to predict the temperature of the water after 3.5 minutes.

thanks

2. Originally Posted by mat
hello some questions for bearings I have to draw these manually.

a man walks 3.6 kilometres on bearing 235 degrees from A to B.
draw a diagram and find how far B is west of A and how far B is south of A?
How to find the angles in the triangle?

...
I assume that you have already drawn the sketch.

The direction exactly to North correspond to a bearing of 0°. E = 90°. S = 180° and W = 270°.

You are dealing with a right triangle whose hypotenuse is known to you. Use Sine and Cosine function to calculate the length of the legs.

3. Originally Posted by mat
...
cartesian plane
draw a diagram of the cartesian plane to show point p (1,3) and point q (-3,-2)
I know how to find gradient, midpoint and length but
draw a line 5x-4y-4=0
and explain why this is parallel to pq.
in the diagram calculate the angle made by line pq and the positive x axis.
1. Calculate the slope between P and Q:

$\displaystyle m_{PQ}=\frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

2. Rewrite the equation of the line:

$\displaystyle 5x-4y-4=0 ~\implies~ 5x-4 = 4y~\implies~\boxed{y = \frac54 x-1}$

3. Since the slopes are equal the line is parallel to PQ.

The angle between the line and the positive x-axis is calculated by the Arcus-Tangens function:

$\displaystyle \alpha = \arctan(m) ~\implies~ \alpha = \arctan \left(\frac54\right) \approx 51.34^\circ$

5. Originally Posted by earboth
Arcus-Tangens function:
Uh...arctangent...?

6. Originally Posted by Mathstud28
Uh...arctangent...?
He's just using the Latin to look fancy: sinus, secans, tangens, etc. Interestingly enough, Wikipedia says that sinus means "breast," and that it was a mistranslation from Arabic. I'm not sure how reliable that is, but breasts do have a sort of sinusoidal shape I suppose, so it works out.

7. Originally Posted by Reckoner
He's just using the Latin to look fancy: sinus, secans, tangens, etc.
I only used the word used in German for the inverse Tangens function. Sorry if I confused you ....
Interestingly enough, Wikipedia says that sinus means "breast," and that it was a mistranslation from Arabic. I'm not sure how reliable that is, but breasts do have a sort of sinusoidal shape I suppose, so it works out.
To my knowledge the word sinus is derived from the Persian word seneš which means bosom which refers to the curve between the breasts.