1. ## 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?

2. Originally Posted by Mr Rayon
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, $\displaystyle \Delta y = 0$

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

$\displaystyle 150^{\circ} T = -60^{\circ}$

$\displaystyle 053^{\circ} T = 37^{\circ}$

$\displaystyle 8\sin(-60) \cdot \left(\frac{3}{4}\right) + v\sin(37) \cdot \left(\frac{4}{3}\right) = 0$

$\displaystyle -8\sin(60) \cdot \left(\frac{3}{4}\right) + v\sin(37) \cdot \left(\frac{4}{3}\right) = 0$

$\displaystyle v\sin(37) \cdot \left(\frac{4}{3}\right) = 8\sin(60) \cdot \left(\frac{3}{4}\right)$

$\displaystyle v = \frac{8\sin(60) \cdot \left(\frac{3}{4}\right)}{\sin(37) \cdot \left(\frac{4}{3}\right)}$

$\displaystyle v \approx 6.5$ km/hr

$\displaystyle d = vt = v \left(\frac{4}{3}\right) \approx 8.6$ km

3. ## Trigonometry

Hello Mr Rayon
Originally Posted by Mr Rayon
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 $\displaystyle A$, the point where he changes direction $\displaystyle B$, and his final position $\displaystyle C$. Then we have:

• $\displaystyle \angle BAC = 150-90 = 60^o = A$

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

• $\displaystyle \angle ABC = 60+53 = 113^o = B$

(i)

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

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

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