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408 lines
16 KiB
HTML
408 lines
16 KiB
HTML
<!DOCTYPE html>
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<html lang="en"><!-- #BeginTemplate "/Templates/Main.dwt" --><!-- DW6 -->
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<!-- Mirrored from www.mathsisfun.com/activity/walk-in-desert-2.html by HTTrack Website Copier/3.x [XR&CO'2014], Sat, 29 Oct 2022 00:58:19 GMT -->
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<title>Activity: A Walk in the Desert 2</title>
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<meta name="Description" content="Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.">
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</header>
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<div class="mid">
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<nav>
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<div id="logo"><a href="../index.html"><img src="../images/style/logo.svg" alt="Math is Fun"></a></div>
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<article id="content" role="main">
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Activity:
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A Walk in the
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Desert 2</h1>
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<p class="center">How to find what <b>direction</b> to travel in</p>
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<h2>Crash!</h2>
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<p style="float:right; margin: 10px;"><img src="images/walk-3.jpg" alt="walk in desert plane" height="310" width="150"></p>
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<p>
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If you haven't met Jade
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yet, then you should do the activity <a href="walk-in-desert.html">A
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Walk
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in the Desert</a> first. </p> <br>
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<br>
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Jade crash-landed in the desert, but came up with a cunning plan to
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find the nearest village:<p></p>
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<ul>
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<div class="bigul">
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<li>Fill up a water
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bottle from the plane, and take a compass,</li>
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<li>Then walk 1 km north, change direction and
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walk 2 km east, then 3 km south, 4 km west, 5 km north, 6 km east,
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and so on, like this:</li>
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</div>
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</ul>
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<p class="center"><img src="images/walk-1.gif" alt="walk1" height="248" width="271"></p>
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<p align="left">This way Jade will find the village no matter what direction it is in, and can (hopefully) find the way back to the plane for fresh water and shade when needed.</p><br>
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<p>But if Jade does <b>not </b>find the village, we need a way to return to the plane
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every few hours for rest and a refill of the water bottle.</p>
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<div class="center80">
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<p>The <b>distances</b> were worked out in <a href="walk-in-desert.html">Activity: A
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Walk
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in the Desert</a></p>
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<p>Now we need to find the <b>directions</b>.</p>
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</div>
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<p>To get back to the plane from point A is easy: retrace the
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steps (go south for 1 km).</p>
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<p>But what about point B? What direction
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should Jade walk from B to get back to the plane?</p>
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<p>We looked at this triangle before:</p>
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<p class="center"><img src="images/walk2.gif" alt="walk2" height="100" width="134"></p>
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<p>and calculated the distance OB = √5 km</p>
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<p>To find the direction we need to calculate an <b>angle</b>, like angle ABO, which is marked
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θ
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in the following diagram:</p>
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<p class="center"><img src="images/walk2-8.gif" alt="walk8" height="99" width="134"></p>
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<p>To find the size of angle θ we need to use <a href="../algebra/trigonometry.html">Trigonometry</a></p>
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<p>We know all three sides, but it's easier to use the whole numbers, so
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we will use the Opposite
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AO = 1 and the Adjacent AB =
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2. <a href="../algebra/sohcahtoa.html">SOHCAHTOA</a> tells us we should use Tangent:</p>
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<div class="so">
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tan(θ) = opposite/adjacent = 1/2 = 0.5</div>
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<p>Now use the <b>tan<sup>-1</sup></b> button or the <b>atan</b> button on your calculator:</p>
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<div class="so">θ = 26.6°</div>
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<p>So, the angle is 26.6°</p>
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<p>But what direction is that?</p>
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<p class="center"><img src="images/walk2-9.jpg" alt="walk9" height="97" width="172"></p>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="../measure/images/compass-rose.svg" alt="compass rose" height="240" width="240"></p>
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<p>Well, it's somewhere between south
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and west, but nearer to west than south. So maybe we could say west
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south-west.</p>
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<p>But that's not very accurate. Jade might miss the plane! Maybe it won't matter too much in this case since B isn't too
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far away from the plane.</p>
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<p>But we
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need to be more accurate for the other points.</p>
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<p> </p>
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<div style="clear:both"></div>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="../measure/images/compass-bearing.svg" alt="compass bearing" height="220" width="219"></p>
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<p>So let's use <b>three-figure bearings</b>.</p>
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<h2>What are
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Three-Figure Bearings?</h2>
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<p>Three-figure bearings are an alternative to compass bearings that are
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much more precise. They are measured in a special way:</p>
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<ul>
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<li>Start measuring from the direction North</li>
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<li>Measure clockwise</li>
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<li>Give the bearing using three figures (or more than three if there's a decimal)</li>
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</ul>
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<p>Airline pilots and ships' helmsmen
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use three-figure
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bearings.</p>
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<div class="example">
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<h3>Examples</h3>
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<p>The four main compass bearings (North, East, South and West)
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are multiples of 90°:</p>
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<p class="center"><img src="../measure/images/three-figure-a.svg" alt="four main compass bearings (North 000, East 090, South 180 and West 270)" height="170" width="475"></p>
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<p>Notice that east, for example is 090° rather than 90°
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because it is given as three figures.</p>
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<p>The advantage of
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three-figure bearings is that they describe any direction uniquely:</p>
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<p class="center"><img src="../measure/images/three-figure-b.svg" alt="three figure bearings examples" height="161" width="327"></p>
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<p>Note that the last one has four figures (three in front of the decimal
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point and one after) but it is still a "three-figure
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bearing", the .4 just gives more accuracy.</p>
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</div>
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<p>Now compare this last example with the direction
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Jade needs to head to get back to the plane at O:</p>
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<p class="center"><img src="images/walk2-13.gif" alt="walk13" height="204" width="375"></p>
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<p>They show the same direction. So how is 243.4° related to the 26.6° angle we obtained before?</p>
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<p>The answer is easy: 270° - 26.6° = 243.4°</p>
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<h2>Your Turn</h2>
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<p>Now you can begin filling out the table below, up to point E (we will use another method for points F to J).</p>
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<p>(Note: distances are calculated in <a href="walk-in-desert.html">A
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Walk
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in the Desert</a>).</p>
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<p>Use a right angled
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triangle to help you to calculate
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the three-figure bearing that Jade needs to walk to head
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back to the plane at O:</p>
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<table align="center" cellspacing="2" cellpadding="2" border="1">
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<tbody>
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<tr>
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<td>Point</td>
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<td>Distance walked<br>
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altogether</td>
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<td>Distance (in a<br>
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straight line)
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from O</td>
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<td>Three-figure bearing<br>
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to head back
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to O</td>
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</tr>
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<tr>
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<td>O</td>
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<td>0</td>
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<td>0</td>
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<td>Not applicable</td>
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</tr>
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<tr>
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<td>A</td>
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<td>1</td>
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<td>1</td>
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<td>180°</td>
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</tr>
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<tr>
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<td>B</td>
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<td>3</td>
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<td>√5</td>
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<td>243.4°</td>
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</tr>
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<tr>
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<td>C</td>
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<td>6</td>
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<td><br></td>
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<td><br></td>
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</tr>
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<tr>
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<td>D</td>
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<td><br></td>
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<td><br></td>
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<td><br></td>
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</tr>
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<tr>
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<td>E</td>
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<td><br></td>
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<td><br></td>
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<td><br></td>
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</tr>
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</tbody>
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</table>
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<h2>Using Polar Coordinates</h2>
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<p>In <a href="walk-in-desert.html">A
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Walk
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in the Desert</a>, <a href="../data/cartesian-coordinates.html">Cartesian Coordinates</a> are used to calculate the
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distance (in a straight line) from O:</p>
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<p class="center"><img src="images/walk-2.gif" alt="walk4" height="273" width="303"></p>
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<p>Using <b>Cartesian Coordinates</b> we mark a point by how far along and how far up:</p>
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<p class="center"><img src="../geometry/images/coordinates-cartesian.svg" alt="cartesian coordinates " height="232" width="348"></p>
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<p class="larger">But there's another kind of coordinates you can use, called <a href="../polar-cartesian-coordinates.html">Polar
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Coordinates</a>.</p>
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<p>Using <b>Polar Coordinates</b> you mark a point by how far away and what angle it is:</p>
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<p class="center"><img src="../geometry/images/coordinates-polar.svg" alt="polar coordinates" height="232" width="348"></p>
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<p>So the point <b>(12, 5)</b> in Cartesian coordinates is the same as the
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point <b>(13, 22.6°)</b> in Polar coordinates.</p>
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<p>That is what we want! A <b>distance</b> and <b>direction</b> for Jade to walk.</p>
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<p>To convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):</p>
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<div class="center80">
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<p class="center larger"><b align="center">r = √( x<sup>2</sup> + y<sup>2 </sup>)</b></p>
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<p class="center larger"><b align="center"><i>θ</i> = tan<sup>-1 </sup>( y / x )</b></p>
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</div>
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<p>Let's do the calculations again for point B. x = 2 and y = 1, so:</p>
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<div class="indent50px">
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<p class="larger"><b align="center">r = √( x<sup>2</sup> + y<sup>2 </sup>)= √( 2<sup>2</sup> + 1<sup>2 </sup>)= √( 4 + 1)= √5</b></p>
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<p class="larger"><b align="center"><i>θ</i> = tan<sup>-1 </sup>( y / x ) = tan<sup>-1 </sup>( 1/2 ) = 26.6°</b></p>
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</div>
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<p class="center"> </p>
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<p class="larger">So the polar coordinates of the point B are (√5, 26.6°)</p>
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<p>But what is the three-figure
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bearing?</p>
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<p> </p>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="../geometry/images/cartesian-quadrants.svg" alt="Quadrants" height="191" width="250"></p>
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<p>Well there is a simple rule based on which <a href="../algebra/trig-four-quadrants.html">Quadrant</a> the point is in:</p>
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<ul>
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<li>For points in Quadrants I, II and III (points B, F, J, E, I, D
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and H), <b>subtract the angle from 270°</b></li>
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</ul>
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<ul>
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<li>For points in Quadrant IV (points C and G), <b>subtract
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the angle from 630°</b> (yes that is <b>630°</b>, not 360°)</li>
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</ul><div style="clear:both"></div>
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<p>So for B (in Quadrant I), θ = 26.6° and the
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three-figure bearing is <b>270° - 26.6° =
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243.4°</b></p>
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Let's try another point:
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<div class="example">
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<h3>For point I, x= -4 and y = 5, so:</h3>
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<div class="indent50px">
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<p class="larger">r = √( x<sup>2</sup> + y<sup>2 </sup>)= √( (-4)<sup>2</sup> + 5<sup>2 </sup>)= √( 16 + 25)= √41</p>
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<p class="larger"><i>θ</i> = tan<sup>-1 </sup>( y / x ) = tan<sup>-1 </sup>( 5/-4 ) = tan<sup>-1 </sup>(-1.25) = 128.7°</p>
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</div>
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<p>Point I is in Quadrant II, so the three-figure bearing is <b>270° - 128.7° = 141.3°</b></p>
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</div>
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<p>Now you should be able to complete the following table:</p>
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<table align="center" cellspacing="2" cellpadding="2" border="1">
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<tbody>
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<tr>
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<td style="text-align:center;">Point</td>
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<td align="center" width="80">Value of r</td>
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<td align="center" width="80">Value of θ</td>
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<td align="center" width="140">Polar coordinate</td>
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<td align="center" width="100">Three-figure bearing<br>
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to head back to O</td>
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</tr>
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<tr>
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<td align="center" height="30">O</td>
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<td style="text-align:center;">0</td>
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<td style="text-align:center;">0°</td>
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<td style="text-align:center;">(0, 0°)</td>
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<td style="text-align:center;">Not applicable</td>
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</tr>
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<tr>
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<td align="center" height="30">A</td>
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<td style="text-align:center;">1</td>
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<td style="text-align:center;">90°</td>
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<td style="text-align:center;">(1, 90°)</td>
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<td style="text-align:center;">180°</td>
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</tr>
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<tr>
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<td align="center" height="30">B</td>
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<td style="text-align:center;">√5</td>
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<td style="text-align:center;">26.6°</td>
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<td style="text-align:center;">(√5, 26.6°)</td>
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<td style="text-align:center;">243.4°</td>
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</tr>
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<td align="center" height="30">C</td>
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<td style="text-align:center;"><br></td>
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<td align="center" height="30">D</td>
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<td align="center" height="30">E</td>
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<td style="text-align:center;"><br></td>
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<td style="text-align:center;"><br></td>
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<td style="text-align:center;"><br></td>
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<td style="text-align:center;"><br></td>
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<tr>
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<td align="center" height="30">F</td>
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<td style="text-align:center;"><br></td>
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<tr>
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<td align="center" height="30">G</td>
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<td style="text-align:center;"><br></td>
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<td style="text-align:center;"><br></td>
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<td style="text-align:center;"><br></td>
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<tr>
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<td align="center" height="30">H</td>
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<td style="text-align:center;"><br></td>
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<td align="center" height="30">I</td>
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<td style="text-align:center;">√41</td>
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<td style="text-align:center;">128.7°</td>
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<td style="text-align:center;">(√41, 128.7°)</td>
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<td style="text-align:center;">141.3°</td>
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<tr>
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<td align="center" height="30">J</td>
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<td style="text-align:center;"><br></td>
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</tbody>
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</table>
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<p> </p>
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<div class="related">
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<a href="walk-in-desert.html">A Walk in the Desert</a>
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<a href="index.html">Activity Index</a>
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