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Math Help - Trig prob

  1. #1
    Member Mr Rayon's Avatar
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    Trig prob

    A cross-country runner runs at 8 km/h on a bearing of 150 T for 45 mins, then changes direction to a bearing of 053 T and runs for 80 mins until he is due east of the starting point.

    a) (i) How far was the second part of the run?
    (ii) What was his speed for this section?
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by Mr Rayon View Post
    A cross-country runner runs at 8 km/h on a bearing of 150 T for 45 mins, then changes direction to a bearing of 053 T and runs for 80 mins until he is due east of the starting point.

    a) (i) How far was the second part of the run?
    (ii) What was his speed for this section?
    note that overall displacement north/south, \Delta y = 0

    directions relative to the x-axis (east) ...

     <br />
150^{\circ} T = -60^{\circ}<br />

     <br />
053^{\circ} T = 37^{\circ}<br />


     <br />
8\sin(-60) \cdot \left(\frac{3}{4}\right) + v\sin(37) \cdot \left(\frac{4}{3}\right) = 0<br />

     <br />
-8\sin(60) \cdot \left(\frac{3}{4}\right) + v\sin(37) \cdot \left(\frac{4}{3}\right) = 0<br />

     <br />
v\sin(37) \cdot \left(\frac{4}{3}\right) = 8\sin(60) \cdot \left(\frac{3}{4}\right)<br />

     <br />
v = \frac{8\sin(60) \cdot \left(\frac{3}{4}\right)}{\sin(37) \cdot \left(\frac{4}{3}\right)}<br />

    v \approx 6.5 km/hr

    d = vt = v \left(\frac{4}{3}\right) \approx 8.6 km
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  3. #3
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    Trigonometry

    Hello Mr Rayon
    Quote Originally Posted by Mr Rayon View Post
    A cross-country runner runs at 8 km/h on a bearing of 150 T for 45 mins, then changes direction to a bearing of 053 T and runs for 80 mins until he is due east of the starting point.

    a) (i) How far was the second part of the run?
    (ii) What was his speed for this section?
    Call the starting point A, the point where he changes direction B, and his final position C. Then we have:

    • \angle BAC = 150-90 = 60^o = A


    • AB = \tfrac{3}{4}\times 8 = 6 km = c


    • \angle ABC = 60+53 = 113^o = B

    (i)

    • Use the angle sum of triangle to calculate \angle C


    • Use the Sine Rule to calculate a (= BC): \frac{a}{\sin A} = \frac{c}{\sin C}

    (ii) This distance takes 80 mins = \frac{80}{60}=\frac{4}{3} hours. So divide the distance BC by this time, \frac{4}{3}, to find his speed in km/h.

    Grandad
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