SOLUTION: At 12noon,a ship steaming steadily due east at 10knots, observed a light house at a bearing 060° and half an hour later the same light house is observed at a bearing of 040° . ca

Question 1166568: At 12noon,a ship steaming steadily due east at 10knots, observed a light house at a bearing 060° and half an hour later the same light house is observed at a bearing of 040° . calculate the time to the nearest minutes at which the bearing will be 020° and the distance from the ship to the light house at this time.[take 1knots=1nautical mile per hour].

Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
This is a navigation problem that can be solved using trigonometry. The ship's movement and the lighthouse form a series of right and non-right triangles.
Here is the step-by-step calculation.
## 🚢 Part 1: Establish Initial Triangle and Speed
* **Ship's Speed ($v$):** 10 knots (10 nautical miles/hour).
* **Initial Observation Time ($t_1$):** 12:00 PM.
* **Second Observation Time ($t_2$):** 12:30 PM (half an hour later).
* **Distance Travelled ($d$):** In 0.5 hours: $d = v \times t = 10 \text{ knots} \times 0.5 \text{ hr} = 5 \text{ nautical miles}$.
Let $L$ be the lighthouse, $P_1$ be the ship's position at 12:00 PM, and $P_2$ be the ship's position at 12:30 PM.
* $P_1 P_2 = 5$ n.m.
* $P_1$ bearing to $L$ is $060^\circ$.
* $P_2$ bearing to $L$ is $040^\circ$.
## 📐 Part 2: Find Angles in Triangle $P_1 P_2 L$
The ship sails **due east** (bearing $090^\circ$). The bearings are measured clockwise from North ($000^\circ$).
1. **Angle $\angle L P_1 P_2$:**
* Since the ship is moving East ($090^\circ$) and the lighthouse is at $060^\circ$, the angle between the ship's path and the line of sight $P_1 L$ is:
$$\angle L P_1 P_2 = 090^\circ - 060^\circ = 30^\circ$$
2. **Angle $\angle P_1 L P_2$:**
* The North lines at $P_1$ and $P_2$ are parallel. The angle between North and $P_1 L$ is $060^\circ$.
* The angle between the North line at $P_1$ and the ship's path $P_1 P_2$ is $090^\circ$.
* Using the alternate interior angles: The angle at $P_1$ between the North line and $P_1 L$ is $60^\circ$.
* The angle at $L$ between $P_1 L$ and the line extending North from $P_2$ (a parallel line) is $60^\circ$ (Alternate Interior Angles).
* The bearing at $P_2$ is $040^\circ$. The angle between the North line at $P_2$ and $P_2 L$ is $40^\circ$.
* Therefore, the angle $\angle P_1 L P_2$ is the difference between the two bearings:
$$\angle P_1 L P_2 = 060^\circ - 040^\circ = 20^\circ$$
3. **Angle $\angle L P_2 P_1$:**
* The sum of angles in $\triangle P_1 P_2 L$ is $180^\circ$.
* $$\angle L P_2 P_1 = 180^\circ - (\angle L P_1 P_2 + \angle P_1 L P_2)$$
* $$\angle L P_2 P_1 = 180^\circ - (30^\circ + 20^\circ) = 130^\circ$$
---
## 🧭 Part 3: Calculate Distance $P_2 L$
We use the **Law of Sines** on $\triangle P_1 P_2 L$ to find the distance $P_2 L$.
$$\frac{P_2 L}{\sin(\angle L P_1 P_2)} = \frac{P_1 P_2}{\sin(\angle P_1 L P_2)}$$
$$\frac{P_2 L}{\sin(30^\circ)} = \frac{5}{\sin(20^\circ)}$$
$$P_2 L = \frac{5 \times \sin(30^\circ)}{\sin(20^\circ)}$$
$$P_2 L = \frac{5 \times 0.5}{0.3420} \approx 7.310 \text{ n.m.}$$
---
## ⏱️ Part 4: Calculate Time to Bearing $020^\circ$
Let $P_3$ be the position where the bearing is $020^\circ$. Since the ship is moving East, $P_3$ will be East of $P_2$.
In $\triangle P_3 L P_2$:
* $\angle L P_3 P_2 = 090^\circ - 020^\circ = 70^\circ$.
* $\angle P_3 L P_2 = 040^\circ - 020^\circ = 20^\circ$.
* $\angle L P_2 P_3 = 180^\circ - (\angle L P_3 P_2 + \angle P_3 L P_2) = 180^\circ - (70^\circ + 20^\circ) = 90^\circ$.
Since $\angle L P_2 P_3 = 90^\circ$, this is a **right-angled triangle**!
### Calculate Distance $P_2 P_3$ (Distance to Travel)
We can use the cosine function in the right-angled $\triangle L P_2 P_3$:
$$\cos(\angle L P_3 P_2) = \frac{P_3 P_2}{L P_3}$$
This is missing $L P_3$. Use tangent:
$$\tan(\angle P_3 L P_2) = \frac{P_2 P_3}{P_2 L}$$
$$P_2 P_3 = P_2 L \times \tan(20^\circ)$$
$$P_2 P_3 = 7.310 \times 0.3640$$
$$P_2 P_3 \approx 2.659 \text{ n.m.}$$
### Calculate Time
Time required to travel $2.659$ n.m. at $10$ knots:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{2.659}{10} = 0.2659 \text{ hours}$$
Convert hours to minutes:
$$\text{Minutes} = 0.2659 \text{ hr} \times 60 \text{ min/hr} \approx 15.95 \text{ minutes}$$
**Time to the nearest minute:** $\mathbf{16 \text{ minutes}}$.
### Calculate Time of Observation
The time is 16 minutes after the second observation at 12:30 PM.
$$\text{Time} = 12:30:00 \text{ PM} + 00:16:00 = \mathbf{12:46 \text{ PM}}$$
---
## 📏 Part 5: Distance to Lighthouse at Time $P_3$
We need to find the distance $L P_3$. In the right-angled $\triangle L P_2 P_3$, we can use the sine function:
$$\sin(\angle P_3 L P_2) = \frac{P_2 P_3}{L P_3}$$
$$L P_3 = \frac{P_2 P_3}{\sin(20^\circ)} \text{ OR } \cos(\angle L P_3 P_2) = \frac{P_3 P_2}{L P_3}$$
Using the cosine relationship:
$$L P_3 = \frac{P_2 P_3}{\cos(70^\circ)}$$
$$L P_3 = \frac{2.659}{0.3420}$$
$$L P_3 \approx \mathbf{7.775 \text{ n.m.}}$$
---
## ✅ Final Answer
| Calculation | Result |
| :---: | :---: |
| Time elapsed from 12:30 PM | 16 minutes |
| **Time of $\mathbf{020^\circ}$ Bearing** | **12:46 PM** |
| **Distance from Ship to Lighthouse** | **7.78 n.m. (or 7.775 n.m.)** |
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