Maths Help: Bearing Problems


Uploaded by vividmaths on 14.05.2012

Transcript:
Hi guys I am going to share with you reveal to you one of the biggest, most powerful,
most sineificant statements I could ever offer you in term of how to do bearing in fact how
to do trigonometry and all of math. This is particularly appropriate and suitable for
when you can have a problem like as follow. Now the drawing of the diagram, explain the
diagram taking those words and showing them as a diagram is the key thing so here is the
statement. A problem well drawn is a problem half solve so if you can draw takes those
words and turn them into wonderful picture you are half way there and the rest of it
is just smooth sailing you can just cruise through the solution so let us take a closer
look at this diagram and how it came about from this word so let us go through this words
again. A ship sails on a bearing of 200 degrees so you will always start from a port that
is quarter start somewhere and it start as a bearing so try the way you do across from
where you start. The first thing you do is find out where the bearing, the bearing is
says 200 degrees so you start zero flip all the way around for 200 hundred degrees that
is past 180 and now it is less than 270 it is going to be here somewhere. you can draw
rough if you want so you draw a line that is the first step now looking at the words
again. Now this is from O is in the question from O there we are there is a line 200 degrees
bearing if P is 18 nautical miles further west, further west than O, this O this west
that will use west yeah. so that is 18 nautical mile so let us put that in, out west so that
is 80, 80 nautical mile at that way. Ok so we got that way it is going at the way. What
other information do we have? Ahófind how far the ship is sail? Another one, is how
far this is? So let us put an X that is our missing pronumeral. So what do we have? We
have bearing 200 degrees let us write that in, we have 80 nautical miles west of O and
we have the missing pronumeral along here the distance that we are looking for X all
the way to the ship, so that is what we were after. Now we got the diagram drawn and displayed
we can now we can start to reduce had actually do this so we go to the step no. 2.
Step no. 2 we are actually looking at the diagram and calculating the angle that we
need because you probably noticed we have a right angle triangle here right? So we want
to find probably this angle in here so that can tell what this line here is, because we
have this line here, we got the 80 nautical miles yeah. we are looking for X here we need
that angle in there so that angle is going to be the angle Q O P, so angle-- that we are after, once we got it looks like it is going to be adjacent over hypotenuse looks like probably going to be
a cos, but for now degrees-- that is in total, all we need to do is take away the bearing all the way around
to here which is 200 degreesótake away 200 that is rather easy 270 ñ 200 is 70 degrees.
so we can safely put that angle in there so let us do that 70 degrees also . Ok now we
have that angle we can work out the trig rusher as I said earlier then determine what exactly
that length along here is. So let us do that now, so it looks like it is going to be --- over
here see how the 80 is next to the seventy? It is going up like that is next and it is
going to be ADJ and down here with this long length here is definitely a hypotenuse is
in it? The long length is not attach to the right angle. Ok now let us actually work with
this triangle now so we can actually work out on that link so it is going to be a cos
rusher because it is adjacent to the hypotenuse so write that down so we have --cos that angle
there which 70 degrees = adjacent which is 80 let us put that in-- / now it is going
to be all over the missing X right there the hypotenuse is in it? so let us put that X
there , ok this is our equation. Cos 70 =80/or an X, we are looking for X so I am going to
multiply both sides by X, times by X and times by X, multiplying both by X we are going to
be left on this side with a X cos 70 degrees = 80.now let us make X the subject let us
bring cos70 underneath and let us divide both sides by cos70 that will yield, X= that is
going to be all over cos 70, and on the top we have 80 of course , there is the 80 comes
there is the cos70 comes there, let me just draw a little arrow here, there, that is there.
Ok on the calculator 80 / cos70 gives you, the length of here the missing pronumeral
right there the X = on the calculator 233.9 newton nautical miles
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