Question 1168158: A cross-country race covers a course which takes shape of a triangle, KLM. Runners are to start at K and run 4 km due north to a point L.
From L, they are to run 5 km on a bearing of 300° to a point, M, and then run directly to K.
(a) draw a diagram to represent the triangular course carefully labeling:
(i) the points K, L and M
(II) the distances 4 km and 5 km
(iii) the north direction
(iv) the bearing 300°
(b) determine, using your diagram:
(i) the distance KM in km
(ii) the bearing of M from K
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **(a) Draw a diagram to represent the triangular course carefully labeling:**
**(i) the points K, L and M**
**(ii) the distances 4 km and 5 km**
**(iii) the north direction**
**(iv) the bearing 300°**
```mermaid
graph TD
K -- 4 km --> L
L -- 5 km, bearing 300° --> M
M --> K
style K fill:#fff
style L fill:#fff
style M fill:#fff
direction LR
subgraph Compass
N[North]
style N fill:#eee
end
L --o N
L -- 60° --> M
```
**Explanation of the diagram:**
* **Points K, L, and M:** These represent the vertices of the triangular course.
* **Distance KL:** L is 4 km due north of K, so the line segment KL is vertical and points upwards.
* **Distance LM:** From L, the runner goes 5 km on a bearing of 300°. A bearing of 300° is measured clockwise from North. So, starting from the North direction at L, we rotate 300° clockwise. This places M in the northwest quadrant relative to L. The angle between the North direction at L and the line segment LM is $360° - 300° = 60°$.
* **North Direction:** A North arrow is shown originating from point L to indicate the reference for the bearing.
* **Bearing 300°:** The angle of 300° is implied by the 60° angle shown relative to the North direction at L.
**(b) determine, using your diagram:**
**(i) the distance KM in km**
**(ii) the bearing of M from K**
To determine the distance KM and the bearing of M from K, we need to use trigonometry on the triangle KLM.
**(i) Distance KM:**
We can use the Law of Cosines on triangle KLM. We know the lengths of sides KL = 4 km and LM = 5 km, and we need to find the length of side KM. To use the Law of Cosines, we need the angle $\angle KLM$.
From the diagram, the North direction at L makes an angle of 90° with the horizontal (East or West). The bearing of LM is 300°, which means the angle between LM and the North direction is 60° (as calculated above). Therefore, the angle $\angle KLM$ is $90° + 60° = 150°$.
Now, using the Law of Cosines:
$KM^2 = KL^2 + LM^2 - 2(KL)(LM) \cos(\angle KLM)$
$KM^2 = 4^2 + 5^2 - 2(4)(5) \cos(150°)$
$KM^2 = 16 + 25 - 40 (-\frac{\sqrt{3}}{2})$
$KM^2 = 41 + 20\sqrt{3}$
$KM^2 \approx 41 + 20(1.732)$
$KM^2 \approx 41 + 34.64$
$KM^2 \approx 75.64$
$KM = \sqrt{75.64} \approx 8.697$
Rounding to a reasonable number of decimal places (e.g., two), the distance KM is approximately **8.70 km**.
**(ii) the bearing of M from K:**
To find the bearing of M from K, we need to find the angle that the line segment KM makes with the North direction at K, measured clockwise. Let this angle be $\theta$.
We can use the Law of Sines on triangle KLM to find the angle $\angle LKM$.
$\frac{\sin(\angle LKM)}{LM} = \frac{\sin(\angle KLM)}{KM}$
$\frac{\sin(\angle LKM)}{5} = \frac{\sin(150°)}{8.697}$
$\sin(\angle LKM) = \frac{5 \times \sin(150°)}{8.697} = \frac{5 \times 0.5}{8.697} = \frac{2.5}{8.697} \approx 0.2874$
$\angle LKM = \arcsin(0.2874) \approx 16.7°$
Since L is due North of K, the line KL is along the North direction. The point M is to the West of the North line from K (because to reach M from L, we moved on a bearing of 300°). Therefore, the bearing of M from K will be $360° - \angle LKM$.
Bearing of M from K $\approx 360° - 16.7° = 343.3°$
Rounding to one decimal place, the bearing of M from K is approximately **343.3°**.
Final Answer: The final answer is $\boxed{a) \text{Diagram as described above}, b) (i) \approx 8.70 \text{ km}, (ii) \approx 343.3°}$
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