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<title>Solving SAS Triangles</title>
<meta name="description" content="SAS means Side, Angle, Side" />
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<h1 align="center">Solving SAS Triangles</h1>
<p align="center"><i>&quot;SAS&quot; means &quot;Side, Angle, Side&quot;</i></p>
<table border="0" align="center">
<tr>
<td><img src="images/triangle-sas.svg" alt="SAS Triangle" /></td>
<td><p>&quot;<i>SAS</i>&quot; is when we know two sides and the angle between them.</p></td>
</tr>
</table>
<p>To solve an SAS triangle</p>
<ul>
<li>use <a href="trig-cosine-law.html">The Law of Cosines</a> to calculate the unknown side, </li>
<li>then use <a href="trig-sine-law.html">The Law of Sines</a> to find the smaller of the other two angles, </li>
<li>and then use <a href="../proof180deg.html">the three angles add to 180&deg;</a> to find the last angle. </li>
</ul>
<div class="example">
<h3>Example 1</h3>
<p style="float:left; margin: 0 10px 5px 0;">
<img src="images/trig-sasex1.gif" alt="trig SAS example" /></p>
<p>In this triangle we know:</p>
<ul>
<li>angle A = 49&deg;</li>
<li>b = 5 </li>
<li>and c = 7 </li>
</ul>
<p>&nbsp;</p>
<p>To solve the triangle we need to find side <b>a</b> and angles <b>B</b> and <b>C</b>.</p>
<p>Use The Law of Cosines to find side <b>a</b> first:</p>
<p class="large">a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> &minus; 2bc cosA</p>
<div class="so">a<sup>2</sup> = 5<sup>2</sup> + 7<sup>2</sup> &minus; 2 &times; 5 &times; 7 &times; cos(49&deg;)</div>
<div class="so">a<sup>2</sup> = 25 + 49 &minus; 70 &times; cos(49&deg;)
</div>
<div class="so">a<sup>2</sup> = 74 &minus; 70 &times; 0.6560...
</div>
<div class="so">a<sup>2</sup> = 74 &minus; 45.924... = 28.075...</div>
<div class="so">a = &radic;28.075...</div>
<div class="so"> a = 5.298... </div>
<div class="so"> a = <b>5.30</b> to 2 decimal places</div>
<p>&nbsp; </p>
<p>Now we use the The Law of Sines to find the smaller of the other two angles.</p>
<p><b>Why the smaller angle?</b> Because the inverse sine function gives answers less than 90&deg; even for angles greater than 90&deg;. By choosing the smaller angle (a triangle won't have two angles greater than 90&deg;) we avoid that problem. Note: the smaller angle is the one facing the shorter side.</p>
<p>&nbsp;</p>
<p>Choose angle B:</p>
<p class="large">sin B / b = sin A / a</p>
<div class="so"> sin B / 5 = sin(49&deg;) / 5.298... </div>
<p>Did you notice that we didn't use <b>a = 5.30</b>. That number is rounded to 2 decimal places. It's much better to use the unrounded number 5.298... which should still be on our calculator from the last calculation.</p>
<div class="so"> sin B = (sin(49&deg;) &times; 5) / 5.298... </div>
<div class="so"> sin B = 0.7122... </div>
<div class="so"> B = sin<sup>&minus;1</sup>(0.7122...)</div>
<div class="so"> B = <b>45.4&deg;</b> to one decimal place</div>
<p>&nbsp;</p>
<p>Now we find angle C, which is easy using 'angles of a triangle add to 180&deg;':</p>
<div class="so">C = 180&deg; &minus; 49&deg; &minus; 45.4&deg;</div>
<div class="so">C = <b>85.6&deg;</b> to one decimal place</div>
<p>&nbsp; </p>
<p>Now we have completely solved the triangle i.e. we have found all its angles and sides.</p>
</div><p>&nbsp;</p>
<div class="example">
<h3>Example 2</h3>
<p style="float:left; margin: 0 10px 5px 0;">
<img src="images/trig-sasex2.gif" width="269" height="120" alt="trig SAS example" /></p>
<p>This is also an SAS triangle.</p>
<div style="clear:both"></div>
<p>First of all we will find <b>r</b> using The Law of Cosines:</p>
<p class="large">r<sup>2</sup> = p<sup>2</sup> + q<sup>2</sup> &minus; 2pq cos R</p>
<div class="so">r<sup>2</sup> = 6.9<sup>2</sup> + 2.6<sup>2</sup> &minus; 2 &times; 6.9 &times; 2.6 &times; cos(117&deg;)
</div>
<div class="so">r<sup>2</sup> = 47.61 + 6.76 &minus; 35.88 &times; cos(117&deg;)
</div>
<div class="so">r<sup>2</sup> = 54.37 &minus; 35.88 &times; (&minus;0.4539...) </div>
<div class="so">r<sup>2</sup> = 54.37 + 16.289...
= 70.659... </div>
<div class="so"> r = &radic;70.659... </div>
<div class="so">r = 8.405...
= <b>8.41</b> to 2 decimal places</div>
<p>&nbsp; </p>
<p>Now for The Law of Sines.</p>
<p>Choose the smaller angle? We don't have to! Angle R is greater than 90&deg;, so angles P and Q must be less than 90&deg;.</p>
<p>&nbsp;</p>
<p class="large">sin P / p = sin R / r</p>
<div class="so"> sin P / 6.9 = sin(117&deg;) / 8.405... </div>
<div class="so"> sin P = ( sin(117&deg;) &times; 6.9 ) / 8.405...</div>
<div class="so"> sin P = 0.7313...</div>
<div class="so"> P = sin<sup>&minus;1</sup>(0.7313...) </div>
<div class="so"> P = <b>47.0&deg;</b> to one decimal place </div>
<p>&nbsp;</p>
<p>Now we will find angle Q using 'angles of a triangle add to 180&deg;':</p>
<div class="so"> Q = 180&deg; &minus; 117&deg; &minus; 47.0&deg;</div>
<div class="so"> Q = <b>16.0&deg;</b> to one decimal place</div>
</div>
<p>Mastering this skill needs lots of practice, so ...</p>
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<a href="trig-solving-triangles.html">Solving Triangles</a>
<a href="trig-solving-practice.html">Triangle Solving Practice</a>
<a href="trig-sine-law.html">The Law of Sines</a>
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