document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166568>Question 1166568</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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].<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852741 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852741\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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.\r<BR>\n" );
document.write( "\n" );
document.write( "Here is the step-by-step calculation. \r<BR>\n" );
document.write( "\n" );
document.write( "## 🚢 Part 1: Establish Initial Triangle and Speed\r<BR>\n" );
document.write( "\n" );
document.write( "* **Ship's Speed ($v$):** 10 knots (10 nautical miles/hour).<BR>\n" );
document.write( "* **Initial Observation Time ($t_1$):** 12:00 PM.<BR>\n" );
document.write( "* **Second Observation Time ($t_2$):** 12:30 PM (half an hour later).<BR>\n" );
document.write( "* **Distance Travelled ($d$):** In 0.5 hours: $d = v \times t = 10 \text{ knots} \times 0.5 \text{ hr} = 5 \text{ nautical miles}$.\r<BR>\n" );
document.write( "\n" );
document.write( "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.\r<BR>\n" );
document.write( "\n" );
document.write( "* $P_1 P_2 = 5$ n.m.<BR>\n" );
document.write( "* $P_1$ bearing to $L$ is $060^\circ$.<BR>\n" );
document.write( "* $P_2$ bearing to $L$ is $040^\circ$.\r<BR>\n" );
document.write( "\n" );
document.write( "## 📐 Part 2: Find Angles in Triangle $P_1 P_2 L$\r<BR>\n" );
document.write( "\n" );
document.write( "The ship sails **due east** (bearing $090^\circ$). The bearings are measured clockwise from North ($000^\circ$).\r<BR>\n" );
document.write( "\n" );
document.write( "1.  **Angle $\angle L P_1 P_2$:**<BR>\n" );
document.write( "    * 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:<BR>\n" );
document.write( "        $$\angle L P_1 P_2 = 090^\circ - 060^\circ = 30^\circ$$\r<BR>\n" );
document.write( "\n" );
document.write( "2.  **Angle $\angle P_1 L P_2$:**<BR>\n" );
document.write( "    * The North lines at $P_1$ and $P_2$ are parallel. The angle between North and $P_1 L$ is $060^\circ$.<BR>\n" );
document.write( "    * The angle between the North line at $P_1$ and the ship's path $P_1 P_2$ is $090^\circ$.<BR>\n" );
document.write( "    * Using the alternate interior angles: The angle at $P_1$ between the North line and $P_1 L$ is $60^\circ$.<BR>\n" );
document.write( "    * 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).<BR>\n" );
document.write( "    * The bearing at $P_2$ is $040^\circ$. The angle between the North line at $P_2$ and $P_2 L$ is $40^\circ$.<BR>\n" );
document.write( "    * Therefore, the angle $\angle P_1 L P_2$ is the difference between the two bearings:<BR>\n" );
document.write( "        $$\angle P_1 L P_2 = 060^\circ - 040^\circ = 20^\circ$$\r<BR>\n" );
document.write( "\n" );
document.write( "3.  **Angle $\angle L P_2 P_1$:**<BR>\n" );
document.write( "    * The sum of angles in $\triangle P_1 P_2 L$ is $180^\circ$.<BR>\n" );
document.write( "    * $$\angle L P_2 P_1 = 180^\circ - (\angle L P_1 P_2 + \angle P_1 L P_2)$$<BR>\n" );
document.write( "    * $$\angle L P_2 P_1 = 180^\circ - (30^\circ + 20^\circ) = 130^\circ$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 🧭 Part 3: Calculate Distance $P_2 L$\r<BR>\n" );
document.write( "\n" );
document.write( "We use the **Law of Sines** on $\triangle P_1 P_2 L$ to find the distance $P_2 L$.\r<BR>\n" );
document.write( "\n" );
document.write( "$$\frac{P_2 L}{\sin(\angle L P_1 P_2)} = \frac{P_1 P_2}{\sin(\angle P_1 L P_2)}$$<BR>\n" );
document.write( "$$\frac{P_2 L}{\sin(30^\circ)} = \frac{5}{\sin(20^\circ)}$$<BR>\n" );
document.write( "$$P_2 L = \frac{5 \times \sin(30^\circ)}{\sin(20^\circ)}$$<BR>\n" );
document.write( "$$P_2 L = \frac{5 \times 0.5}{0.3420} \approx 7.310 \text{ n.m.}$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## ⏱️ Part 4: Calculate Time to Bearing $020^\circ$\r<BR>\n" );
document.write( "\n" );
document.write( "Let $P_3$ be the position where the bearing is $020^\circ$. Since the ship is moving East, $P_3$ will be East of $P_2$.\r<BR>\n" );
document.write( "\n" );
document.write( "In $\triangle P_3 L P_2$:<BR>\n" );
document.write( "* $\angle L P_3 P_2 = 090^\circ - 020^\circ = 70^\circ$.<BR>\n" );
document.write( "* $\angle P_3 L P_2 = 040^\circ - 020^\circ = 20^\circ$.<BR>\n" );
document.write( "* $\angle L P_2 P_3 = 180^\circ - (\angle L P_3 P_2 + \angle P_3 L P_2) = 180^\circ - (70^\circ + 20^\circ) = 90^\circ$.\r<BR>\n" );
document.write( "\n" );
document.write( "Since $\angle L P_2 P_3 = 90^\circ$, this is a **right-angled triangle**!\r<BR>\n" );
document.write( "\n" );
document.write( "### Calculate Distance $P_2 P_3$ (Distance to Travel)\r<BR>\n" );
document.write( "\n" );
document.write( "We can use the cosine function in the right-angled $\triangle L P_2 P_3$:<BR>\n" );
document.write( "$$\cos(\angle L P_3 P_2) = \frac{P_3 P_2}{L P_3}$$<BR>\n" );
document.write( "This is missing $L P_3$. Use tangent:<BR>\n" );
document.write( "$$\tan(\angle P_3 L P_2) = \frac{P_2 P_3}{P_2 L}$$<BR>\n" );
document.write( "$$P_2 P_3 = P_2 L \times \tan(20^\circ)$$<BR>\n" );
document.write( "$$P_2 P_3 = 7.310 \times 0.3640$$<BR>\n" );
document.write( "$$P_2 P_3 \approx 2.659 \text{ n.m.}$$\r<BR>\n" );
document.write( "\n" );
document.write( "### Calculate Time\r<BR>\n" );
document.write( "\n" );
document.write( "Time required to travel $2.659$ n.m. at $10$ knots:<BR>\n" );
document.write( "$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{2.659}{10} = 0.2659 \text{ hours}$$\r<BR>\n" );
document.write( "\n" );
document.write( "Convert hours to minutes:<BR>\n" );
document.write( "$$\text{Minutes} = 0.2659 \text{ hr} \times 60 \text{ min/hr} \approx 15.95 \text{ minutes}$$\r<BR>\n" );
document.write( "\n" );
document.write( "**Time to the nearest minute:** $\mathbf{16 \text{ minutes}}$.\r<BR>\n" );
document.write( "\n" );
document.write( "### Calculate Time of Observation\r<BR>\n" );
document.write( "\n" );
document.write( "The time is 16 minutes after the second observation at 12:30 PM.<BR>\n" );
document.write( "$$\text{Time} = 12:30:00 \text{ PM} + 00:16:00 = \mathbf{12:46 \text{ PM}}$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 📏 Part 5: Distance to Lighthouse at Time $P_3$\r<BR>\n" );
document.write( "\n" );
document.write( "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:<BR>\n" );
document.write( "$$\sin(\angle P_3 L P_2) = \frac{P_2 P_3}{L P_3}$$<BR>\n" );
document.write( "$$L P_3 = \frac{P_2 P_3}{\sin(20^\circ)} \text{ OR } \cos(\angle L P_3 P_2) = \frac{P_3 P_2}{L P_3}$$<BR>\n" );
document.write( "Using the cosine relationship:<BR>\n" );
document.write( "$$L P_3 = \frac{P_2 P_3}{\cos(70^\circ)}$$<BR>\n" );
document.write( "$$L P_3 = \frac{2.659}{0.3420}$$<BR>\n" );
document.write( "$$L P_3 \approx \mathbf{7.775 \text{ n.m.}}$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## ✅ Final Answer\r<BR>\n" );
document.write( "\n" );
document.write( "| Calculation | Result |<BR>\n" );
document.write( "| :---: | :---: |<BR>\n" );
document.write( "| Time elapsed from 12:30 PM | 16 minutes |<BR>\n" );
document.write( "| **Time of $\mathbf{020^\circ}$ Bearing** | **12:46 PM** |<BR>\n" );
document.write( "| **Distance from Ship to Lighthouse** | **7.78 n.m. (or 7.775 n.m.)** | </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );