Question 1024625: Ok so, the question is more bearings, but there wasn't a category for that. Sorry. But the question is;
A triathlete leaves the start of a race and runs 12km on a bearing of 30 degrees T. She then cycles for 15km on a bearing of 60 degrees T. What are the distance and bearing of her starting point from her position at the end of the cycle leg?
I have tried a few different ways, but I cannot work it out.
Thanks
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! bearing is number of degrees from north.
i think i found the answer.
it was a little complex to get, but i'm pretty sure it's good.
i'm not sure if there's an easier way to get it.
perhaps there is, but i don't know it.
we start with the first leg.
it's a bearing of 30 degrees true (i believe t means true).
that forms a right triangle with 12 km as the hypotenuse as shown in diagram 1.
using trigonometry, the vertical side leg is equal to 12 * sin(60) = 10.39230485 = y1 and the horizontal side leg is equal to 12 * cos(60) = 6 = x1.
we then move on to the second leg.
this forms a right triangle with 15 km as the hypotenuse as shown in diagram 2.
using trigonometry, the vertical side leg is equal to 15 * sin(30) = 7.5 = y2 and the horizontal side leg is equal to 15 * cos(30) = 12.9903038106 = x2.
we get:
x1 = 6
y1 = 10.39230485
x2 = 12.9903038106
y2 = 7.5
now we put the diagrams together and we get diagram 3.
the second leg starts where the first leg finished as shown in the diagram.
the first leg is AB = 12
the second leg is BD = 15.
the two smaller triangles formed are ABC and BDE.
there is a larger triangle formed that is ADF.
we have already found the following distances.
AC = x1 = 6
BC = y1 = 10.39230485
BE = x2 = 12.9903038106
DE = y2 = 7.5
since CF is the same length at BE, then the length of AF is equal to x1 + x2.
the length of AF is therefore equal to 18.99038106
since EF is the same length as BCD, then the length of DF is equal to y1 + y1.
the length of DF is therefore equal to 17.89230485.
since AD is the hypotenuse of right triangle ADF, then the length of AD is equal to the square root of the length of AF squared plus the length of DF squared.
this becomes:
AD = sqrt(18.99038106^2 + 17.89230485^2).
this makes AD equal to a length of 26.09155314.
angle A in triangle ADF is therefore equal to arcsin(DF/AD).
this makes angle A equal to arcsin(17.89230485/26.09155314).
this makes angle A = 43.29468619.
the bearing at point A is therefore equal to 90 - 43.29468619 which is equal to 46.70531381 degrees.
if i did this correctly, your solution should be:
distance from start of first leg to end of second leg = 26.09155314 km.
bearing from start of first leg to end of second leg = 46.70531381 degrees.
the diagrams are shown below:
check it out.
i think it's good but always leave open the possibility that i might be wrong.
if the answer agrees with what the solutuion should be then you're good.
if not, let me know what the solution is supposed to be and i'll give it another crack.
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