2357
/www/nrich/html/content/id/2357/
1
2011-02-01T00:00:01
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<ul id="stemLinks">
<li><a href="http://nrich.maths.org/2844">Warm-up</a></li>
<li><a href="http://nrich.maths.org/6683">Try this next</a></li>
<li><a href="http://nrich.maths.org/7484">Think higher</a></li>
<li><a href="http://nrich.maths.org/6843">Read: mathematics</a></li>
<li><a href="http://plus.maths.org/issue51/package/MMP_navigation_toolkit.pdf">Read: science</a></li>
<li><a href="http://plus.maths.org/content/os/issue55/package/index">Explore further</a></li>
</ul>
<div> </div>
<p style="">An observer is on top of a lighthouse which is 25 metres high. The radius of the earth is 6367 kilometres.</p>
<p>How far from the foot of the lighthouse, measured on the surface of the earth, is the horizon that the observer can see?</p>
<p align="center"><mdo:image align="top" alt="Picture of the white cliffs of dover" bgcolor="" height="156" src="4-3White_Cliffs_of_Dover.jpg" width="125"></mdo:image></p>
<p style="clear: right;"> </p>
<p>It is 32 kilometres (21 miles) across the Channel between England and France and you can see France from the cliffs of Dover.</p>
<p>How high must the cliffs be for it to be possible to see that distance?</p></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0">
<br></br>
<p>The solution below is based on the one sent in by Barinder of Langley Grammar School. We had quite a number of correct, well laid out solutions to this problem this month including those from Roy of Allerton High School, Dan (no address given) and Calum of Wayland High School. <mdo:image alt="cross sectional diagram of earth and lighthouse" height="259" src="far-h1.gif" width="220"></mdo:image><br></br>
Although definitely not in proportion, this makes the problem seem a lot easier. The question is asking for the length of the arc I have coloured red. To get this, I decided to find the angle $ \theta $ on the diagram, and use the equation<br></br>
<br></br>
Arc Length $ = \frac {\theta}{360} \times 2 \pi r $ where $ \theta $ is measured in degrees and r is the radius.</p>
<p>$\angle OAB = 90^\circ$, since it is where a tangent and a radius of a circle meet - it is a circle theorem.<br></br>
<br></br>
Thus, the triangle AOB can be drawn as follows:<br></br>
<br></br>
<mdo:image alt="Righta-,nagled triangle showing relevant lengths" height="222" src="far-h2.gif" width="211"></mdo:image><br></br>
<br></br>
<br></br>
We can now use trigonometry to find $ \theta $:<br></br>
<br></br>
$$\begin{align*} \cos \theta &= \frac {6367000}{6367025} \\ \cos \theta &= 0.99999607 \\ \theta &= \cos^{-1} (0.99999607) = 0.1606^\circ \mbox{(4 d.p.)}\end{align*}$$<br></br>
<br></br>
Substitute this into the equation for the arc length of a circle earlier to obtain the length required:<br></br>
<br></br>
Arc Length $= \frac {0.1606}{360} \times 2\pi r. $<br></br>
<br></br>
Arc Length $= 0.000446 \times 2 \times \pi \times 6367000 = 17,842.3m = 17.8 km $<br></br>
<br></br>
For this next part, we are given the arc length, since this corresponds to the distance between England and France. The diagram is therefore:</p>
<div><mdo:image alt="Cross section of the Earth and the White Cliffs of dover" height="335" src="far-h3.gif" width="326"></mdo:image></div>
<div>This is essentially the reverse of the previous question. We need to find the angle $\alpha$ first, and to do this, we consider the arc length of the sector OAD of the circle:</div>
<br></br>
<div>Arc Length $ = \frac {\alpha}{360} \times 2 \pi r$.</div>
<br></br>
<div>So $32000 = \frac {\alpha}{360} \times 2 \times \pi \times 6367000$.</div>
<br></br>
<div>Then $32000 \times 360 = \alpha \times 2 \times \pi \times 6367000 $.</div>
<br></br>
<div>So $ \alpha = 0.288^\circ \mbox{(3 d.p.)}$</div>
<br></br>
<div><mdo:image alt="Right-angled triangle showing lengths" height="222" src="far-h4.png" width="211"></mdo:image></div>
<div>Since we now have the angle $\alpha$, we can consider the triangle AOB:</div>
<br></br>
<div>$ \cos \alpha = \frac {6367000}{6367000 + h} $.</div>
<br></br>
<div>So $ 6367000 + h = \frac {6367000}{\cos (0.288)} $.</div>
<br></br>
<div>So $ 6367000 + h = 6367080.415 $</div>
<br></br>
<div>$ h = 6367080.415 - 6367000 = 80.415 $ m high.</div>
<br></br>
<div>Thus, the cliffs of Dover are $80.4$ metres high.</div>
<br></br>
</mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0">It is not envisaged that <a href="http://nrich.maths.org/2357">this problem</a> would be used as a class problem. It is more appropriate for an enthusiastic student or small group of students looking for a challenge to work on independently.</mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"> <p>
The observer's line of vision is along a tangent to the surface of the
earth at the far point she can see on the horizon. Imagine a right
angled triangle with the observer at one vertex, the far horizon point
at another vertex and the centre of the earth at the third vertex.
</p>
<p>
You have enough information to find the angles in this triangle and
hence the distance required.
</p>
</mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<div class="framework">The solution below is based on the one sent in by Barinder of Langley Grammar School. We had quite a number of correct, well laid out solutions to this problem this month including the ones Roy of Allerton High School and Dan (no address given) and Calum of Wayland High School.</div>
<br></br>
<br></br>
<mdo:image alt="cross sectional diagram of earth and lighthouse" height="259" src="far-h1.gif" width="220"></mdo:image><br></br>
Although definitely not in proportion, this makes the problem seem a lot easier. The question is asking for the length of the arc I have coloured red. To get this, I decided to find the angle $ \theta $ on the diagram, and use the equation<br></br>
<br></br>
Arc Length $ = \frac {\theta}{360} \times 2 \pi r $ where $ \theta $ is measured in degrees and r is the radius $ \angle OAB = 90\deg $, since it is where a tangent and a radius of a circle meet - it is a circle theorem.<br></br>
<br></br>
Thus, the triangle AOB can be drawn as follows:<br></br>
<mdo:image alt="Righta-,nagled triangle showing relevant lengths" height="222" src="far-h2.gif" width="211"></mdo:image><br></br>
<br></br>
We can now use trigonometry to find $ \theta $:<br></br>
<br></br>
$ \cos \theta = \frac {6367000}{6367025} $ $ \cos \theta = 0.99999607 $ $\theta = \cos^{-1} (0.99999607) = 0.1606 \deg $(4.d.p)<br></br>
<br></br>
Substitute this into the equation for the arc length of a circle earlier to obtain the length required:<br></br>
<br></br>
Arc Length $= \frac {0.1606}{360} \times 2\pi r $<br></br>
<br></br>
Arc Length $= 0.000446 \times 2 \times \pi \times 6367000 = 17,842.3m = 17.8 km $<br></br>
<br></br>
For this next part, we are given the arc length, since this corresponds to the distance between England and France. The diagram is therefore:<br></br>
<div><mdo:image alt="Cross section of the Earth and the White Cliffs of dover" height="335" src="far-h3.gif" width="326"></mdo:image></div>
<div>This is essentially the reverse of the previous question. We need to find the angle a first, and to do this, we consider the arc length of the sector OAD of the circle: Arc Length $ = \frac {\alpha}{360} \times 2 \pi r $ So $32,000 = \frac {\alpha}{360} \times 2 \times \pi \times 6367000$ $32,000 \times 360 = \alpha \times 2 \times \pi \times 6367000 $ So $ \alpha = 0.288^o(3.d.p)$ Since we
now have the angle $ \alpha $ , we can consider the triangle AOB:</div>
<div><mdo:image alt="Right-angled triangle showing lengths" height="222" src="far-h4.png" width="211"></mdo:image></div>
<div>Since we now have the angle a, we can consider the triangle AOB:</div>
<br></br>
<div>$ \cos \alpha = \frac {6367000}{6367000 + h} $</div>
<br></br>
<div>So $ 6367000 + h = \frac {6367000}{\cos (0.288)} $</div>
<br></br>
<div>So $ 6367000 + h = 6367080.415 $</div>
<br></br>
<div>$ h = 6367080.415 - 6367000 = 80.415 $ m high</div>
<br></br>
<div>Thus, the cliffs of Dover are 80.4 metres high.</div>
<br></br>
<br></br></mdoxml>
2
5
0
0
0
1
0
Far Horizon
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
Trigonometry
Cosine
Measures and Mensuration
Arcs, sectors and segments
Using, Applying and Reasoning about Mathematics
Real world
2D Geometry, Shape and Space
Tangents
2D Geometry, Shape and Space
Radius (radii) & diameters
Information and Communications Technology
smartphone
Admin
No Teachers Notes
Applications
Maths Supporting SET
Admin
Individual
Applications
STEM - physical world
Secondary Mapping Document
Trigonometry
Secondary Mapping Document
DisplayCabinet