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      <text>5.1.1 (Right-angled triangles)</text>
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      <text><![CDATA[<h1>Right-angled triangles</h1>
<p>Most of us are familiar with the definition of the trigonometry as the ratio of the sides of a right angled triangle, as can been seen in the following diagram.</p>
<p><img style="display: block; margin-left: auto; margin-right: auto;" src="@@PLUGINFILE@@/Right%20angled%20triangle%20trig.png" alt="diagram showing trig as ratios of right angled triangle" width="249" height="192" /></p>
<p>In this case we define the trigonometric functions as the ratio of the lengths of the sides of the triangle as follows: \[ \sin \theta = \frac{O}{H} \qquad \cos \theta = \frac{A}{H} \qquad \tan \theta \frac{O}{A}, \] from which which get the well-remembered mnemonic of SOHCAHTOA, which helps us to remember these definitions. One advantage of these definitions, is that we can use trigonometric functions in order to calculate the area of triangles.</p>
<p>Recall that the area of a triangle is given by the formula Area = base x height. However, if we do not know the perpendicular height of a given triangle, we can use trigonometry to calculate the area. In order to do this we need to know the lengths of two sides of the triangle and the <em>included</em> angle between them. Recall that for a triangle ABC, the angles at vertices \(A,B\) and \(C\) are labelled \(A,B\) and \(C\) respectively. The sides opposite these angles are labelled \(a, b\) and \(c\) as in the following diagram.</p>
<p><img style="display: block; margin-left: auto; margin-right: auto;" src="@@PLUGINFILE@@/Triangle%20labelling.png" alt="diagram showing convention for labelling angles and lengths of a triangle" width="358" height="202" /></p>
<p>Now, note that any triangle that is not right angled must either be acute or obtuse as can be seen in the following diagrams.</p>
<p><img style="display: block; margin-left: auto; margin-right: auto;" src="@@PLUGINFILE@@/Triangle%20area.png" alt="diagram showing obtuse and acute triangle" width="800" height="300" /></p>
<p>In both cases, we define \(h\) to be constructed from the points \(A\) to \(D\). Let us consider the acute case first. Considering triangle ACD this is a right-angled triangle and therefore we can apply the trig definitions above to write: \[ \sin C = \frac{h}{b} \implies h = b \sin C. \] In the obtuse case consider triangle ACD again, to write \[ \sin (180^\circ - C) = \frac{h}{b} \implies h = b\sin C, \] where we have used the identity \(\sin(180^\circ - C) = \sin C\) from the previous section.</p>
<p>Hence, in both cases we have that \(h = b \sin C\). However, we know that the area of a triangle is given by \(1/2 \cdot b \cdot h\), so substituting the first equation into the second we get that the formula for the area of a triangle is \[ \mbox{Area} = \frac{1}{2} a b \sin C. \] Notice that in relation to angle \(C\) here sides \(a\) and \(b\) are the sides which have \(C\) as an included angle. In fact, there is a certain symmetry at play in this equation because using similar techniques we can also see that \[ \mbox{Area} = \frac{1}{2} bc \sin A \qquad \mbox{and} \qquad \mbox{Area} = \frac{1}{2} ac \sin B. \]</p>]]></text>
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