hurricanemk1c 195 Posted November 19, 2013 Report Share Posted November 19, 2013 Inspired by Jack's thread, here's one that had us all thinking at school. Shouldn't be too hard...... Enjoy! Link to post Share on other sites
ddavid 149 Posted November 19, 2013 Report Share Posted November 19, 2013 Ouch, Chuck - do we all really have to go back to skule?!? Now, let me see, mmm - let h = the side opposite the hippopotomus... Cheers - Dai. Link to post Share on other sites
allardjd 1,853 Posted November 20, 2013 Report Share Posted November 20, 2013 i: R to HP = ~47.02 km ii: HP = ~ 164.16 km b; RT = ~ 34.72 km I timed it - it took 16 minutes. Hope it's correct. John Link to post Share on other sites
Aircraft Aviation 2 Posted November 20, 2013 Report Share Posted November 20, 2013 Either I am being stupid or there is something wrong here (most probably the latter). Using the sine rule, letting x be the unknown angle in the bottom right of the triangle: [sin(36)/110] = [sin(x)/80] Solving this for Sin(x), we get: Sin(x) = [sin(36)/110]*[80] Therefore x = Sin^(-1)[0.42748....] Therefore x = 25.3 degrees. This would mean that the total sum of angles in the triangle would be 36 + 124 + 25.3 = 185.3 degrees Clearly this is not correct, on what seems like a basic question!!! Link to post Share on other sites
allardjd 1,853 Posted November 20, 2013 Report Share Posted November 20, 2013 You're on to something, Jack. The third, unlabeled angle must be 20 degrees, of course. I didn't use the Law of Sines (not directly anyway) for my solution so the dichotomy didn't become apparent. One of the distances, 80 or 110 km, or the angles given, 36, 124 and (by deduction) 20, must be incorrect. It's not possible to know which one, of course, but the triangle described by the problem statement is impossible. I don't think applying spherical geometry methods would make enough difference to matter either - the distances are just too short for that to have any appreciable effect. The problem is poorly stated, particularly the statement, "...turning through an angle of 124 [degrees]...". The ship does not turn through 124 degrees. It's origninal heading is 54 degrees, using standard navigational nomenclature. The new course after the turn is 110 degrees, a turn of only 56 degrees. Looking at the diagram it's obvious the turn is much less than 90 degrees, so sloppy work there on the part of whoever framed the problem. Good work smelling the rat. I was dancing on the corpse without having a hint that the problem as stated is flawed. John Link to post Share on other sites
Quickmarch 488 Posted November 21, 2013 Report Share Posted November 21, 2013 Hmmm...HP=193.92, given the conditions they state. 124 degrees is an "included" angle. Correctly stated by John, the actual turn is 180-124=56 degrees, yielding a heading of 110 degrees. Using the problem wording, the ship would sail 80 Km at (90-36)=54 degrees. A subsequent right hand turn of 124 degrees would yield a heading of 178 degrees, or pretty much due south. The correct answer is: You can't get there from here. Link to post Share on other sites
hurricanemk1c 195 Posted November 21, 2013 Author Report Share Posted November 21, 2013 And Jack and John (March) are right - it is physically impossible to create such a triangle. The worrying thing is, this is off last year's exam from Project Maths, something that's been going on for two or three years and *should* make it easier! We are all hoping this doesn't happen this year, when we are doing it! Link to post Share on other sites
Quickmarch 488 Posted November 21, 2013 Report Share Posted November 21, 2013 What is 'Project Maths", Kieren? The question reminds me very much of my experience in high school. There were way too many of these silly problems that seemed to do nothing more than force the student to arrive at a "correct" answer by the approved method (geometry "proofs" come to mind). Lord save us from the student that tended to think a little outside the box. In this case, it would be interesting to see how many students used the obvious 20 degrees for the third angle and simply did the trig to arrive at the correct answers. The easiest way to show this is graphically: Note that the second (inner) triangle matches JA's calculations. Link to post Share on other sites
hurricanemk1c 195 Posted November 22, 2013 Author Report Share Posted November 22, 2013 Project Maths is a new way of teaching Maths, supposedly to make it more relatable to today's world. Most of the questions hide what they are asking you (it seems anyway) Info here - http://www.projectmaths.ie/ I think they accepted any answer as long as some calculations either proved it or not Link to post Share on other sites
J G 927 Posted November 22, 2013 Report Share Posted November 22, 2013 I think they accepted any answer as long as some calculations either proved it or not Oh great! now maths has gone PC on us! So all those big red crosses on my maths book must have traumatized me. I think I will sue the school for mental cruelty. Link to post Share on other sites
Christopher Low 63 Posted November 24, 2013 Report Share Posted November 24, 2013 Is the poor bugger in the ship still trying to find Port P? Link to post Share on other sites
hurricanemk1c 195 Posted November 24, 2013 Author Report Share Posted November 24, 2013 Probably! Link to post Share on other sites
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