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  <resource>
  <id>5614</id>
  <path>/www/nrich/html/content/id/5614/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
 
&lt;table border=&quot;0&quot;&gt;
&lt;tbody&gt;
&lt;tr&gt;
&lt;td&gt;&lt;mdo:image height=&quot;380&quot; width=&quot;553&quot; alt=&quot;house&quot; src=&quot;house.JPG&quot;&gt;&lt;/mdo:image&gt;&lt;/td&gt;
&lt;td&gt;
&lt;div&gt;The diagram shows a cross-sectional view of a house.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;

&lt;div&gt;The roof space is 12 metres across and 1.5 metres high at the
highest point in the centre.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div&gt;Find the angle of slope of the roof.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;&lt;/td&gt;
&lt;/tr&gt;
&lt;/tbody&gt;
&lt;/table&gt;
&lt;div&gt;Also find the sloping distance $b$ from the top of the roof to
the top of the walls.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;

&lt;div&gt;The hottest part of the day is when the sun is directly
overhead. The roof overhangs by a distance $a$ and it shades all
the windows from the sun while the sun's rays are inclined at 8
degrees or less to the vertical. The lowest part of the windows is
one metre above the ground and the angle $\phi$ in the diagram is 8
degrees. Find the length of the overhang, $a$.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;We have had a number of excellent solutions to this problem from Year 11 pupils in Sharnbrook School in Bedforshire. And they didn&amp;#39;t all solve the problem in the same way...&lt;/span&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;Jennie, Emily, Peter and Hannah sent in this solution, and used Pythagoras&amp;#39; Theorem to find one of their answers.&lt;/span&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;mdo:image src=&quot;Raising%20the%20Roof%201.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;Megan, Shivani, Will, Austen and Michael managed to find &lt;em&gt;b&lt;/em&gt; in a different way, using what they knew about the angles in the roof.&lt;/span&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;mdo:image src=&quot;Raising%20the%20Roof%202.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;Mollie, Meg and Jack also used Pythagoras, and relabeled the angles and sides to make it easier for them to answer the questions. Perhaps they should have returned to the original labels at the end of their solution...&lt;/span&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;mdo:image src=&quot;Raising%20the%20Roof%203.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;Olivia, Alex and Toby sent in very clear solutions to the first and third parts of the problem. We&amp;#39;ll assume that they could have also done the second part!&lt;/span&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;mdo:image src=&quot;Raising%20the%20Roof%204.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;Just simple trigonometry.&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</noteXML>
  <clueXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;Just simple trigonometry.&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</clueXML>
  <canonXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
Slope of roof (pitch) = $\tan^{-1}(1.5/6)= 0.244978663$ radians =
14.03624347 degrees.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$b= 6/\cos 14.04^o = 6.184658438$ = 6185 mm to the nearest
mm.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
To calculate $a$ first find the angles in the triangle formed by
the overhang, the wall of the house and the rays of the sun. These
angles in degrees are: $75.96375653, 8, 96.03624347$ ($
1.325817664, 0.13962634, 1.67614865$ radians). Using the Sine Rule
$$a= {5.5 \sin 8\over \sin 96} = 0.769719708 = 770 {\rm mm}$$ to
the nearest mm. &lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</canonXML>
  <end_user_role>2</end_user_role>
  <difficulty>4</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>0</keystage3>
  <keystage4>1</keystage4>
  <keystage4plus>0</keystage4plus>
  <title>Raising The Roof</title>
  <description>How far should the roof overhang to shade windows from the mid-day sun?</description>
  <spec_group>Trigonometry
    <specifier>Sine rule</specifier>
  </spec_group>
  <spec_group>Trigonometry
    <specifier>Mixed trig ratios</specifier>
  </spec_group>
</resource>