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QCE Stuff => QCE Mathematics Subjects => QCE Subjects + Help => QCE Mathematical Methods => Topic started by: Luke_8064 on November 16, 2019, 12:13:46 pm

Title: Trigonometry II Worded Problem HELP!
Post by: Luke_8064 on November 16, 2019, 12:13:46 pm
Hi all,

I was just wondering if someone could help me understand the question below (and potentially run through part b for me) :
In a cross-country run, a competitor runs at an average speed of 10m/minute for 20 minutes along a track from the starting point (A) ,directly north to checkpoint B. From checkpoint B they make their way across to checkpoint C, then back to point A. The distance between C and A is 250 m. The bearing of point C from point A has been recorded as 068°T.

(a) Draw a diagram to show the course.
(b) How far is the first leg of the run (i.e. From A to B).
(c) Write down the value of angle A.
(d) Find the distance between the checkpoints B and C.
(e) Determine the bearing of C from B to the nearest degree.

Thanks,
Luke.

PS: Sorry if I am posting in the wrong subject section. I just thought as I am in Foundation Mathematics Methods, I would just post this question here. 
Title: Re: Trigonometry II Worded Problem HELP!
Post by: Bri MT on November 16, 2019, 12:24:26 pm
Hey Luke,

Have you been able to draw the diagram for this question? The most crucial part of understanding the wording in the question is recognising the track as a triangle.

For part b in particular, they want you to find the distance using distance = speed x time.
In this case, the units line up nicely (think: metres/minute x minutes = metres) so you don't have to do any annoying conversions.

Hope this helps :)
Title: Re: Trigonometry II Worded Problem HELP!
Post by: Luke_8064 on November 16, 2019, 05:13:37 pm
Hey Luke,

Have you been able to draw the diagram for this question? The most crucial part of understanding the wording in the question is recognising the track as a triangle.

For part b in particular, they want you to find the distance using distance = speed x time.
In this case, the units line up nicely (think: metres/minute x minutes = metres) so you don't have to do any annoying conversions.

Hope this helps :)

Hi Bri,

Yes, I did try to draw a diagram however I got confused where the bearing had to go for checkpoint C and A. When I first drew the diagram, I put the 68 degrees at angel C but I thought this was wrong and changed it to angle A. This lead me to be even more confused because in part C of this question it says to declare the what the value of angle A is equal to. This is where I am really confuzzled.

If you could help me understand this part of the problem I would sure appreciate it!  ;D
Luke.
Title: Re: Trigonometry II Worded Problem HELP!
Post by: RuiAce on November 17, 2019, 10:16:32 am
Hi Bri,

Yes, I did try to draw a diagram however I got confused where the bearing had to go for checkpoint C and A. When I first drew the diagram, I put the 68 degrees at angel C but I thought this was wrong and changed it to angle A. This lead me to be even more confused because in part C of this question it says to declare the what the value of angle A is equal to. This is where I am really confuzzled.

If you could help me understand this part of the problem I would sure appreciate it!  ;D
Luke.
The bearing of the point \(C\) from the point \(A\) is \(68^\circ\). This means that if you draw the northerly direction from \(C\), the point \(A\) is inclined at \(68^\circ\) relative to it. (So your second attempt was the correct one there.)

This is not the same thing as angle \(A\). Angle \(A\) denotes \(\angle BAC\) (or alternatively \(\angle CAB\)), which is one of the angles in the actual triangle. This is not the same thing as a bearing.