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      <text>4.4.1.1 Applications of trigonometry to triangles</text>
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      <text><![CDATA[<h3>Applications of trigonometry to triangles</h3>
<p><br></p>
<h5>Area of a triangle</h5>

<p>The area \(S\) of any triangle is given by \(S=\dfrac{1}{2}\times (\text{base}) \times (\text{perpendicular} \, \text{height})\) where 'perpendicular height' means the perpendicular distance from the side called the 'base' to the opposite vertex. Thus
    for the right-angled triangle shown bellow \(\quad S=\dfrac{1}{2} \ b\ a.\) For the obtuse-angled triangle (b) the area is \(S=\dfrac{1}{2}\ b \ h.\)</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/TrigFig20.svg" alt="(a) Image of a right-angled triangle (b) Image of an odtuse-angled triangle." width="391" height="127" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>If we use \(C\) to denote the angle \(ACB\) in Figure (b) then</p>
<p>\(\qquad \sin (180-C) = \dfrac{h}{a} \qquad\) (triangle \(BCD\) is right-angled)</p>
<p>Therefore,</p>
<p>\( \qquad h=a \sin (180-C) = a\sin C \quad\) (see the graph of the sine wave or expand \(\sin (180-c)\))</p>
<p>\(\qquad \boxed{S=\dfrac{1}{2}\,b\,a \sin C} \qquad \qquad \qquad \qquad\) 1(a)</p>
<p>By other similar constructions we could demonstrate that</p>
<p>\(\qquad \boxed{S=\dfrac{1}{2}\, a\, c\, \sin B } \qquad \qquad \qquad \qquad\) 1(b)</p>
<p>and</p>
<p>\(\qquad \boxed{S=\dfrac{1}{2}\, b\, c\, \sin A} \qquad \qquad \qquad \qquad\) 1(c)</p>
<p>Note the pattern here: in each formula for the area the angle involved is the one <strong>between</strong> the sides whose lengths occur in that expression.</p>
<p>Clearly if \(C\) is a right-angle (so \(\sin C=1\)) then</p>
<p>\(\qquad S=\dfrac{1}{2}\, b\, a \quad\) as for Figure (a).</p>
<p><strong>Note</strong>: from now on we will not generally write ‘\(\equiv\)’ but use the more usual ‘\(=\)’.</p>


<p><br></p>
<h5>The Sine rule</h5>
<p>The Sine rule is a formula which, if we are given certain information about a triangle, enables us to fully 'solve the triangle' i.e. obtain the lengths of all three sides and the value of all three angles.</p>
<p>To show the rule we note that from the formulae (1a) and (1b) for the area \(S\) of the triangle \(ABC\) in Figure 33 we have</p>
<p>\(\qquad ba\sin C=a c\sin B \quad\) or \(\quad \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)</p>
<p>Similarly using (1b) and (1c)</p>
<p>\(\qquad a c\sin B = bc\sin A \quad\) or \(\quad \dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)</p>

<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p></p>
    <center><strong>The Sine Rule</strong></center>
    <p></p>
    <p>For <strong>any</strong> triangle \(ABC\) where \(a\) is the length of the side opposite angle \(A\), \(b\) the side length opposite angle \(B\) and \(c\) the side length opposite angle \(C\) states</p>
    \[\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\]
</div>

<p><br></p>
<h5>Use of the Sine rule</h5>

<p>To be able to fully determine all the angles and sides of a triangle it follows from the Sine rule that we must know</p>
<p> \(\qquad\) <strong>either</strong> \(\quad\) two angles and one side: (knowing two angles of a triangle really means that all three are known since the sum of the angles is \(180^\circ\)</p>
<p>\(\qquad\) <strong>or</strong> \(\quad\) two sides and an angle <strong>opposite</strong> one of those two sides.</p>

<p><br></p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Solve the triangle \(ABC\) given that \(a=32\, \text{cm},\, \, b=46\, \text{cm}\, \) and angle \( B=63.25^\circ.\)</p>

<h4 class="HELM_solution">Solution</h4>
<p>Using the first pair of equations in the Sine rule (previous Key Point) we have</p>
\[ \dfrac{32}{\sin A} = \dfrac{46}{\sin 63.25^\circ}\qquad \therefore\qquad \sin A=\dfrac{32}{46}\sin 63.25^\circ=0.6212\]
<p>so</p>
\[ A= \sin^{-1}(0.6212)=38.4^\circ \quad (\text{by} \, \text{calculator})\]

<p>You should, however, note carefully that because of the form of the graph of the sine function there are <strong>two</strong> angles between \(0^\circ\) and \(180^\circ\) which have the same value for their sine i.e. \(x\) and \((180-x).\) See figure bellow.</p>

<figure class="HELM_figure">
  <img src="@@PLUGINFILE@@/TrigFig22.svg" alt="Graph of a sine function" width="300" height="194" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>In our example</p>
<p>\( \qquad A=\sin^{-1}(0.6212) = 38.4^\circ\)</p>
<p>or</p>
<p>\(\qquad A=180^\circ-38.4^\circ = 141.6^\circ.\)</p>
<p>However since we are given that angle \(B\) is \(63.25^\circ\), the value of \(141.6^\circ\) for angle \(A\) is clearly impossible.</p>
<p>To complete the problem we simply note that</p>
<p>\( \qquad C=180^\circ-(38.4^\circ+63.25^\circ) = 78.35^\circ \)</p>
<p>The remaining side \(c\) is calculated from the Sine rule, using either \(a\) and \(\sin A\) or \(b\) and \(\sin B.\)</p>]]></text>
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