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		<title>Understanding the Law of Sines</title>
		<link>https://complete-concrete-concise.com/mathematics/understanding-the-law-of-sines/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Wed, 13 Mar 2013 11:25:10 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[sine]]></category>
		<category><![CDATA[sine law]]></category>
		<category><![CDATA[trigonometry]]></category>
		<category><![CDATA[understanding]]></category>
		<guid isPermaLink="false">http://complete-concrete-concise.com/?p=2651</guid>

					<description><![CDATA[<p>While trigonometry is used to solve problems involving right angle triangles, it can also be applied to triangles that are not right angle triangles. Assumes you understand basic trigonometric concepts. A tutorial on understanding sine, cosine and tangent can be found here. A tutorial on understanding the trigonometric functions and the unit circle can be [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-the-law-of-sines/">Understanding the Law of Sines</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
<p>While trigonometry is used to solve problems involving right angle triangles, it can also be applied to triangles that are not right angle triangles.</p>
<p>Assumes you understand basic trigonometric concepts.</p>
<p>A tutorial on understanding sine, cosine and tangent can be found <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">here</a>.</p>
<p>A tutorial on understanding the trigonometric functions and the unit circle can be found <a href="//complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced">here</a>.</p>
</div>
<h1>What is the Law of Sines</h1>
<p>One of the fundamental properties of two triangles that have the same shape (i.e., they have the same angles) is that the ratio of any two sides is identical &#8211; regardless of the size of the triangles.</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-3.png" alt="" border="0" /></p>
<pre><code>a   A         a   A         b   B
— = —   and   — = —   and   — = —
b   B         c   C         c   C</code></pre>
<p>This property allows you to study a more conveniently sized triangle to learn things about an inconveniently large or small triangle (for example, a triangle formed by the Earth, moon and sun, or a triangle of forces formed by the interaction of atoms or molecules).</p>
<p>The trigonometric functions were developed for right angle triangles because they made it easier to solve problems.</p>
<p>Surprisingly, the trigonometric function <code>sine</code> can be applied to any type of triangle &#8211; provided you know how.</p>
<p>Trigonometry is about ratios and relationships and the <span style="text-decoration: underline;">Law of Sines</span> is about ratios, specifically, <span class="i1">the ratio of a side to the <code>sine</code> of the angle opposite that side is the <span style="text-decoration: underline;">same</span> for all the sides and angles</span>:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2013/03/understanding-sine-law-1.png" alt="" border="0" /></p>
<h1>Example</h1>
<div class="c1">
<div class="c3">
<p>Consider a tree which grows at an angle of 80° relative to the ground. If the measured angle to the top of the tree is 30° and the distance from the measured angle to the base of the tree is 100m, how tall is the tree? How high does the tree stand?</p>
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2013/03/understanding-sine-law-5.png" alt="" border="0" /></p>
<p>Using the <span style="text-decoration: underline;">Law of Sines</span>, we know that the following relations are true:</p>
<pre><code>   A           B           C
———————  =  ———————  =  ———————
sin (a)     sin (b)     sin (c)</code></pre>
<p>Since the height of the tree is <code>A</code>, we are only interested in the following relationship:</p>
<pre><code>   A           C
———————  =  ———————
sin (a)     sin (c)</code></pre>
<p>We know the angles in a triangle sum to 180°, so we know <code>c = 180° - 80° - 30° = 70°</code>.</p>
<p>Substituting into our Sine Law relation we get:</p>
<pre><code>   A           100
————————  =  ————————
sin (30)     sin (70)</code></pre>
<p>Cross multiplying we get:</p>
<pre><code>      100 × sin (30)
A  =  ——————————————
         sin (70)</code></pre>
<p>Solving, we get a height of 53.2m for the tree.</p>
<p>To solve how high the tree stands, we solve it using both standard trigonometry and the <u>Law of Sines</u> to show they give the same answer:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2013/03/understanding-sine-law-6.png" alt="" border="0" /></p>
<h3>Using standard trigonometry:</h3>
<ul>
<li>Drop a perpendicular line from the peak. This is the the height the tree stands. I call it <code>h</code>.</li>
<li>Use the definition for <code>sine</code>: <code>sinθ = opposite / hypotenuse</code></li>
<li>Rearrange the definition of <code>sine</code>: <code>opposite = hypotenuse × sinθ</code></li>
<li>Plug in the known values and solve: <code>opposite = 53.2 × sin(80°) = 52.4m</code></li>
</ul>
<h3>Using the Law of Sines</h3>
<p>Examining the triangle from a <u>Law of Sines</u> point of view, we see:</p>
<ul>
<li>The length of the hypotenuse is 53.2m and the angle opposite it is 90° &#8211; this gives us one ratio</li>
<li>While the length of <code>h</code> is unknow, we do know the angle opposite is 80° &#8211; this gives us the other ratio</li>
</ul>
<p>Writing the ratios together we get:</p>
<pre><code>   h           53.4
————————  =  ————————
sin (80)     sin (90)</code></pre>
<p>When we solve this through, we get the same answer as using standard trigonometry.</p>
</div>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-the-law-of-sines/">Understanding the Law of Sines</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
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		<title>Understanding Basic Trigonometric Identities</title>
		<link>https://complete-concrete-concise.com/mathematics/understanding-basic-trigonometric-identities/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Fri, 01 Mar 2013 10:37:37 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[cos]]></category>
		<category><![CDATA[cosecant]]></category>
		<category><![CDATA[cosine]]></category>
		<category><![CDATA[cot]]></category>
		<category><![CDATA[cotangent]]></category>
		<category><![CDATA[csc]]></category>
		<category><![CDATA[identities]]></category>
		<category><![CDATA[identity]]></category>
		<category><![CDATA[sec]]></category>
		<category><![CDATA[secant]]></category>
		<category><![CDATA[sin]]></category>
		<category><![CDATA[sine]]></category>
		<category><![CDATA[tan]]></category>
		<category><![CDATA[tangent]]></category>
		<category><![CDATA[trigonometric]]></category>
		<category><![CDATA[trigonometry]]></category>
		<category><![CDATA[understanding]]></category>
		<guid isPermaLink="false">http://complete-concrete-concise.com/?p=2631</guid>

					<description><![CDATA[<p>This tutorial assumes you are familiar with the trigonometric functions and their derivation from the unit circle. A tutorial on the trigonometric functions can be found here. A tutorial on the trigonometric functions and the unit circle can be found here. What is an Identity? An identity (in mathematics) is something that is true (more [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-basic-trigonometric-identities/">Understanding Basic Trigonometric Identities</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
<p>This tutorial assumes you are familiar with the trigonometric functions and their derivation from the unit circle.</p>
<p>A tutorial on the trigonometric functions can be found <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">here</a>.</p>
<p>A tutorial on the trigonometric functions and the unit circle can be found <a href="//complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced">here</a>.</p>
</div>
<h1>What is an Identity?</h1>
<p>An identity (in mathematics) is something that is true (more precisely: <em>a tautological relationship</em>).</p>
<p>For example, the following is an identity:</p>
<pre><code>1 + 1 = 2</code></pre>
<p>It says that <code>1 + 1</code> is <u>exactly</u> the same as <code>2</code>.</p>
<p>A common (and important) trigonometric identity is:</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code></pre>
<p>All this identity says is that what is on the <em>left hand side</em> is <u>identical</u> to what is on the <em>right hand side</em>.</p>
<h1>Why Learn / Use Identities?</h1>
<p>Identities (in any branch of mathematics) help us to:</p>
<ul>
<li>solve</li>
<li>simplify</li>
<li>or gain insight into</li>
</ul>
<p>mathematical problems.</p>
<p>Identities are a lot like synonyms in language. Instead of using the word <em>red</em> to describe the colour of something we could use one of the following: <em>carmine, cherry, copper, crimson, ruby, ruddy, rust, scarlet</em>. In language, synonyms give spice to communication; in mathematics, it is used to look at things in a different way.</p>
<p>For example, if you were solving an trigonometric problem and ended up with the following expression:</p>
<pre><code>cos<sup>2</sup>&theta; × tan&theta; × sec&theta; × csc&theta;</code></pre>
<p>you might not think you had achieved much. However, by using basic trigonometric identities, you would see (or maybe have to work out) that the entire expression can be reduced to the value of <code>1</code>.</p>
<h1>How Many Trigonometric Identities are There?</h1>
<p>There are an infinite number of trigonometric identities. Only a few are useful, the remainder generally ending up as problems and exercises for students to solve.</p>
<h1>Reference Image</h1>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-0.png" alt="" border="0" class="centered"/></p>
<div class="c1">
<p>Basic trigonometric identities fall into four groups:</p>
<ol>
<li>Reciprocals</li>
<li>Ratios</li>
<li>Products</li>
<li>Pythagorean Theorem</li>
</ol>
</div>
<h1>Reciprocal Identities</h1>
<p>These identities derive from the fundamental definitions of the trigonometric functions:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-1.png" alt="" border="0" class="centered"/></p>
<p>Given a right angle triangle, we know that:</p>
<pre><code>sin&theta; = A / C</code></pre>
<p>We also know that:</p>
<pre><code>csc&theta; = C / A</code></pre>
<p>It should be obvious that <code>sin&theta; = 1 / csc&theta;</code> because the ratio <code>C / A</code> (cosecant) is the reciprocal of the ratio <code>A / C</code> (sine).</p>
<p>The same reasoning applies to all the <u>reciprocal</u> identities.</p>
<h1>Ratio Identities</h1>
<p>These identities also derive from the fundamental definitions of trigonometric functions and are very similar to the <u>reciprocal</u> identities, except that they consist of the ratio of two trigonometric functions:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-2.png" alt="" border="0" class="centered"/></p>
<p>The most important of these is probably <code>tan&theta; = sin&theta; / cos&theta;</code> (highlighted in blue).</p>
<p>We know that</p>
<pre><code>tan&theta; = A / B     (from a right angle triangle)
tan&theta; = y / x     (from the unit circle)</code></pre>
<p>We also know that <code>tan&theta;</code> is the slope of the hypotenuse.</p>
<p>From the unit circle we know that:</p>
<pre><code>sin&theta; = y
cos&theta; = x</code></pre>
<div class="c2">
<p>It is correct to say that <code>sin&theta; = y</code> is an <u>identity</u> because both are equivalent or identical.</p>
</div>
<p>Using the identities <code>sin&theta;</code> and <code>cos&theta;</code> to substitute for <code>y</code> and <code>x</code>, we get the following:</p>
<pre><code>tan&theta; = y / x = sin&theta; / cos&theta;</code></pre>
<p>A similar derivation can be done using the definitions from a right angle triangle:</p>
<pre><code>sin&theta; = A / C
cos&theta; = B / C</code></pre>
<p>In order to be able to substitute our identities, we need to multiply both the numerator and denominator by <code>1 / C</code></p>
<div class="c1">
<p>One of the principles of algebra is that as long as you multiply or divide ALL the terms by the same amount, the equation remains the same)</p>
</div>
<pre><code>       A   (1 / C)     (A / C)     sin&theta;
tan&theta; = - × &mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  &mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  &mdash;&mdash;&mdash;&mdash;
       B   (1 / C)     (B / C)     cos&theta;</code></pre>
<p>The same sort of reasoning applies to the derivation of the remaining <u>ratio</u> identities.</p>
<div class="c3">
<p>Another class of <em>obvious</em> <u>ratio</u> identities is the ratio of the term to itself:</p>
<pre><code>sin&theta;
&mdash;&mdash;&mdash;&mdash;  =  1
sin&theta;</code></pre>
<p>In all cases, the answer is <code>1</code>.</p>
</p></div>
<h1>Product Identities</h1>
<p>While these can be derived from the fundamental trigonometric identities, it is often easier to use the <u>ratio</u> identities.</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-3.png" alt="" border="0" class="centered"/></p>
<p>Taking <code>cos&theta; × tan&theta; = sin&theta;</code>, we can derive it as follows:</p>
<p><strong>1) Find</strong> an identity for <code>tan&theta;</code>:</p>
<pre><code>       sin&theta;
tan&theta; = &mdash;&mdash;&mdash;&mdash;
       cos&theta;</code></pre>
<p><strong>2) Substitute</strong> this <em>identity</em> for <code>tan&theta;</code> into the identity <code>cos&theta; × tan&theta; = sin&theta;</code>:</p>
<pre><code>       sin&theta;
cos&theta; × &mdash;&mdash;&mdash;&mdash;  =  sin&theta;
       cos&theta;</code></pre>
<p><strong>3) Cancel</strong> the <code>cos&theta;</code> terms:</p>
<pre><code>       sin&theta;
<s>cos&theta;</s> × &mdash;&mdash;&mdash;&mdash;  =  sin&theta;
       <s>cos&theta;</s></code></pre>
<p>The same sort of reasoning applies to the derivation of the remaining <u>product</u> identities.</p>
<div class="c3">
<p>Another class of <em>obvious</em> <u>product</u> identities is the product of the term to its <u>reciprocal</u> identity:</p>
<pre><code>sin&theta; × csc&theta; = 1</code></pre>
<p>In all cases, the answer is <code>1</code>.</p>
</p></div>
<h1>Pythagorean Theorem Identities</h1>
<p>Because the trigonometric functions are derived from a right angle triangle and because the trigonometric functions make use of the sides of a right angle triangle, they can be used in the Pythagorean Theorem:</p>
<pre><code>A<sup>2</sup> + B<sup>2</sup> = C<sup>2</sup></code></pre>
<h3>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</h3>
<p>This is the most common (or most famous) example of a Pythagorean Theorem identity using trigonometric functions.</p>
<p>It can be arranged in many different ways, the three most common would be:</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1
sin<sup>2</sup>&theta; = 1 - cos<sup>2</sup>&theta;
cos<sup>2</sup>&theta; = 1 - sin<sup>2</sup>&theta;</code></pre>
<p>Looking at a right angle triangle, we see that it has 3 sides, normally labelled A, B, and C. With C being the hypotenuse.</p>
<p>Using these sides in the Pythagorean Theorem, we get:</p>
<pre><code>A<sup>2</sup> + B<sup>2</sup> = C<sup>2</sup></code></pre>
<p>To get it to look like <code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code>, we see that we need to get <code>C<sup>2</sup></code> to be equal to <code>1</code>. This can be done by dividing the entire equation by <code>C<sup>2</sup></code>:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>    C<sup>2</sup>
&mdash;  +  &mdash;  =  &mdash;
C<sup>2</sup>    C<sup>2</sup>    C<sup>2</sup></code></pre>
<p>Which gives us:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>
&mdash;  +  &mdash;  =  1
C<sup>2</sup>    C<sup>2</sup></code></pre>
<p>This can be rewritten as:</p>
<pre><code>A   A     B   B
&mdash; &times; &mdash;  <strong>+</strong>  &mdash; &times; &mdash;  =  1
C   C     C   C</code></pre>
<p>We know that <code>A/C = sin&theta;</code> and <code>B/C = cos&theta;</code>, so we can substitute them in the equation giving us:</p>
<pre><code>sin&theta; &times; sin&theta;  +  cos&theta; &times; cos&theta;  =  1</code></pre>
<p>Which, when we collect the terms, gives us the familiar:</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code></pre>
<h3>1 + cot<sup>2</sup>&theta; = csc<sup>2</sup>&theta;</h3>
<p>Just as we obtained <code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code> by dividing the terms in the Pythagorean Theorem by <code>C<sup>2</sup></code>, we can divide all the terms in the Pythagorean Theorem by <code>A<sup>2</sup></code>, which gives us:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>    C<sup>2</sup>
&mdash;  +  &mdash;  =  &mdash;
A<sup>2</sup>    A<sup>2</sup>    A<sup>2</sup></code></pre>
<p>By using similar steps as above, we get the identity:</p>
<pre><code>1 + cot<sup>2</sup>&theta; = csc<sup>2</sup>&theta;</code></pre>
<p>Three more common arrangements would be:</p>
<pre><code>cot<sup>2</sup>&theta; = csc<sup>2</sup>&theta; - 1
csc<sup>2</sup>&theta; = cot<sup>2</sup>&theta; + 1
csc<sup>2</sup>&theta; - cot<sup>2</sup>&theta; = 1</code></pre>
<h3>tan<sup>2</sup>&theta; + 1 = sec<sup>2</sup>&theta;</h3>
<p>Just as we obtained <code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code> by dividing the terms in the Pythagorean Theorem by <code>C<sup>2</sup></code>, we can divide all the terms in the Pythagorean Theorem by <code>B<sup>2</sup></code>, which gives us:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>    C<sup>2</sup>
&mdash;  +  &mdash;  =  &mdash;
B<sup>2</sup>    B<sup>2</sup>    B<sup>2</sup></code></pre>
<p>By using similar steps as above, we get the identity:</p>
<pre><code>tan<sup>2</sup>&theta; + 1 = sec<sup>2</sup>&theta;</code></pre>
<p>Three common arrangements would be:</p>
<pre><code>tan<sup>2</sup>&theta; = sec<sup>2</sup>&theta; - 1
sec<sup>2</sup>&theta; = tan<sup>2</sup>&theta; + 1
sec<sup>2</sup>&theta; - tan<sup>2</sup>&theta; = 1</code></pre>

<h1>Trigonometric Identity Problems</h1>
<p>All of the following are trigonometric identities. Feel free to verify / prove using the identities described in the tutorial:</p>
<div class="c1">
<p>Note, I have used different terms to indicate the angles. This is because the actual term used does not matter.</p></div>
<pre><code>  csc<sup>2</sup>&theta;
&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  cot<sup>2</sup>&theta;
tan<sup>2</sup>&theta; + 1</code></pre>
<pre><code>(1 - sin<sup>2</sup>&alpha;)(1 + tan<sup>2</sup>&alpha;) = 1</code></pre>
<pre><code>sec<sup>2</sup>&phi; - 1
&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  sin<sup>2</sup>&phi;
  sec<sup>2</sup>&phi;</code></pre>
<pre><code>sin&beta; + cos&beta;cot&beta; = csc&beta;</code></pre>
<pre><code>1 - sin<sup>4</sup>&gamma; - 2sin<sup>2</sup>&gamma;cos<sup>2</sup>&gamma; = cos<sup>4</sup>&gamma;</code></pre>
<p>It should be obvious that you can create as many identities as you like. It should also be obvious that most (all?) of these are not as useful as the fundamental identities described in this tutorial.</p>
<div class="c4">
<p>Is the following an identity?</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&beta; = 1</code></pre>
<p>It is an identity <u><strong><em>if and only if</em></strong></u> &theta; = &beta;, otherwise, it is not an identity.</p>
</div>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-basic-trigonometric-identities/">Understanding Basic Trigonometric Identities</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
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		<title>How to Solve Trigonometric Problems</title>
		<link>https://complete-concrete-concise.com/mathematics/how-to-solve-trigonometric-problems/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Thu, 15 Nov 2012 13:45:51 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[cosecant]]></category>
		<category><![CDATA[cosine]]></category>
		<category><![CDATA[cotangent]]></category>
		<category><![CDATA[how to]]></category>
		<category><![CDATA[secant]]></category>
		<category><![CDATA[sine]]></category>
		<category><![CDATA[solving]]></category>
		<category><![CDATA[tangent]]></category>
		<category><![CDATA[trigonometric functions]]></category>
		<category><![CDATA[trigonometry]]></category>
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					<description><![CDATA[<p>This tutorial offers advice on how to solve trigonometric problems and provides several problems worked through in detail. It assumes you are familiar with the trigonometric functions sine, cosine, tangent, secant, cosecant, and cotangent. A basic tutorial can be found here. A more advanced tutorial can be found here. If an explanation / walkthrough is [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/how-to-solve-trigonometric-problems/">How to Solve Trigonometric Problems</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
This tutorial offers advice on how to solve trigonometric problems and provides several problems worked through in detail.<br />
It assumes you are familiar with the trigonometric functions sine, cosine, tangent, secant, cosecant, and cotangent. A basic tutorial can be found <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">here</a>. A more advanced tutorial can be found <a href="//complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced">here</a>.
</div>
<div class="c2">
If an explanation / walkthrough is not clear, please let me know in a comment and I will try to improve the answer.<br />
Identify what it is that is not clear &#8211; just saying &#8220;I didn&#8217;t understand it&#8221; isn&#8217;t very helpful.
</div>
<div class="c1">
If you have a problem you would like to see solved, leave a comment. <strong>Note:</strong> (1) I don&#8217;t do homework, (2) I don&#8217;t promise I will solve it, (3) I don&#8217;t promise to solve it quickly.
</div>
<h1>Don&#8217;t Panic</h1>
<ol>
<li>Don&#8217;t panic or freak out. No one is trying to trick you or trip you up (although, often times, word problems are poorly worded)</li>
<li>Read and understand the problem.</li>
<li>Trigonometric problems are always about right angle triangles. So the trick is to try and break the problem down into right angle triangles.</li>
<li>Trigonometry is about the ratios of the lengths of two sides. If we know the length of two sides of a right angle triangle we can calculate the ratio between those two sides. Those ratios are given special names: sine, cosine, tangent, cosecant, secant, cotangent. Because of the properties of right angle triangles, if we know one of the angles then we also know the ratios of all the sides.</li>
<li>Look for what ratios or partial ratios you can find. Then work from there.</li>
<li>Similar triangles are triangles that have the same shape that have the same angles inside. Often times, we don&#8217;t solve the problem directly but look for a similar triangle. The triangle may be bigger or smaller, but it has the same shape as the triangle we are studying.</li>
</ol>
<h3>Basic Use of Trigonometric Definitions</h3>
<div class="c1">
<div class="c3">
Consider the following right angle triangle having sides of length 3, 4, and 5 units and two angles called alpha and beta. What is sin&alpha;, cos&alpha;, tan&alpha;, sin&beta;, cos&beta;, tan&beta; and the angles &alpha; and &beta;?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/3-4-5-triangle.png" alt="" border="0" /><br />
This problem is about understanding the definition of the trigonometric functions sine, cosine, and tangent.<br />
The triangle is displayed in a familiar way and we can easily calculate the trigonometric relations for &alpha; (remember sin&alpha; is shorthand for &#8220;<em>given a right angle triangle containing an angle &alpha;, what is the ratio of the side opposite the angle to the hypotenuse</em>&#8220;):<br />
We know that <code>sin&alpha; = opposite ÷ hypotenuse</code>, which is <code>3 / 5</code> which is <code>0.6</code>.<br />
We know that <code>cos&beta; = adjacent ÷ hypotenuse</code>, which is <code>4 / 5</code> which is <code>0.8</code>.<br />
We know that <code>tan&alpha; = opposite ÷ adjacent</code>, which is <code>3 / 4</code> which is <code>0.75</code>.<br />
Calculating the trigonometric relations for &beta; is no different, although, we need to correctly identify the opposite and adjacent sides:<br />
We know that <code>sin&beta; = opposite ÷ hypotenuse</code>, which is <code>4 / 5</code> which is <code>0.8</code>.<br />
We know that <code>cos&beta; = adjacent ÷ hypotenuse</code>, which is <code>3 / 5</code> which is <code>0.6</code>.<br />
We know that <code>tan&beta; = opposite ÷ adjacent</code>, which is <code>4 / 3</code> which is <code>1.333...</code><br />
We don&#8217;t know what the angles &alpha; and &beta; are, but we can find out. We know that <code>sin&alpha; = 0.6</code> and <code>sin&beta; = 0.8</code>. All we need to do is find out what angles &alpha; and &beta; have a sine corresponding to 0.6 and 0.8, respectively.<br />
Using the inverse sin function (sin<sup>-1</sup>) on a calculator, we see that &alpha; must be 36.87° for sin&alpha; to be 0.6 and &beta; must be 53.13° for sin&beta; to be 0.8.</p>
<div class="c2">
Before calculator use was wide spread, you would look up the value in a sine table. You can see a sine table with a resolution of 1 degree over <a href="http://www.grc.nasa.gov/WWW/k-12/airplane/tablsin.html">here</a>.
</div>
</div>
<h3>Basic Use and Application of Trigonometric Definitions and Properties of a Right Angle Triangle</h3>
<div class="c2">
<div class="c3">
Given a triangle having sides of length 1, 2, and 3 units, is this a right angle triangle?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/1-2-3-triangle.png" alt="" border="0" /><br />
The simplest way is to see if the Pythagorean Theorem true &#8211; if this is a right angle triangle, we know that: <code>A<sup>2</sup> + B<sup>2</sup> = C<sup>2</sup></code><br />
<code>1<sup>2</sup> + 2<sup>2</sup> = 3<sup>2</sup></code><br />
<code>1 + 4 = 9</code> &#8211; the left hand side and right hand side are not the same, so the triangle is not a right angle triangle<br />
Using trigonometric functions, we can determine the angles &alpha; and &beta; and then check that all the angles sum up to 180°: <code>&alpha; + &beta; + 90° = 180°</code><br />
sin&alpha; = opposite ÷ hypotenuse = 1 / 3 = 0.333&#8230;<br />
sin&beta; = opposite ÷ hypotenuse = 2 / 3 = 0.666&#8230;<br />
Using the inverse sin function (sin<sup>-1</sup>) on a calculator, we determine that &alpha; is 19.47° and &beta; is 41.81°.<br />
Consequently, the sum of the angles is:<br />
<code>19.47° + 41.81° + 90° = 180°</code><br />
<code>151.28° = 180°</code> &#8211; the left hand side and right hand side are not the same, so the triangle is not a right angle triangle</p>
<div class="c1">
The only thing that might seem strange in this example are expressions of the form:<br />
<code>4 = 9</code><br />
These types of expressions are perfectly normal and valid in mathematics. This is because the symbol <code>=</code> does not have the same meaning we are usually used to: in mathematics, <code>=</code> means <u>is the same as</u>.<br />
Written in English, the expression <code>4 = 9</code> would be &#8220;<em>The value of 4 <u>is the same as</u> the value of 9</em>&#8220;. This statement is <u>false</u>. Don&#8217;t get confused and think you have just proven that 4 is the same as 9 because it isn&#8217;t.<br />
In this example, we are saying:</p>
<ol>
<li>The angles in a triangle sum up to 180°</li>
<li>Since this is supposed to be a right angle triangle, we know one of the angles is 90°</li>
<li>Since this is supposed to be a right angle triangle, we can look up the angles &alpha; and &beta; once we know the ratios of the sides</li>
<li>If the sum of all the angles is 180° then we know this is a right angle triangle</li>
<li>If the sum of all the angles is not 180° then we know this is not a right angle triangle.</li>
</ol>
</div>
</div>
<h3>Calculating the Height of a Tree</h3>
<div class="c1">
Potiphar&#8217;s out walking his cat. He notices that he casts a shadow 2 m long. He also notices that a tree in the garden casts a shadow 7 m long. If Potiphar is 1.5 m tall how tall is the tree? What is the angle of elevation of the sun?<br />
<img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/potiphar-1.png" alt="" border="0" /><br />
This is a basic trigonometric problem involving similar triangles. Potiphar and his shadow form a right angle triangle. The tree and its shadow form another right angle triangle. The rays of the sun (which are invisible) form the hypotenuse of these right angle triangles.<br />
<img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/potiphar-2.png" alt="" border="0" /><br />
We know that similar triangles have the same shape and the same angles. We know that the angle or slope of the hypotenuse is the same for both the <u>tree-triangle</u> and the <u>Potiphar-triangle</u> (because rays of sunlight are parallel to each other). Since both triangles are similar, we also know that the ratios of their sides must be the same.<br />
We know the lengths of the sides of the <u>Potiphar-triangle</u> and can use them to calculate a ratio:<br />
<code>A / B = 1.5 / 2 = 0.75</code><br />
Of the tree-triangle, we know the length of one of the sides but not of the other. However, we know that the ratio of the unknown length to the known length (i.e. shadow) must be the same as for the Potiphar-triangle:<br />
<code> a / b = A / B = 0.75</code><br />
Plugging in the values we know (and removing the A / B for clarity):<br />
<code> a / 7 = 0.75</code><br />
Rearranging we get:<br />
<code>a = 0.75 × 7 = 5.25</code><br />
The tree is 5.25 m tall.<br />
To calculate the elevation of the sun, we know the ratio of height to length. In trigonometric terms, this is the definition of tangent.<br />
<code>tan&alpha; = opposite ÷ adjacent = A / B = a / b = 0.75</code><br />
Looking up (or using the inverse key on the calculator) we see that for tan&alpha; to equal 0.75, &alpha; must be 36.87°.</p>
<div class="c2">
Before calculator use was wide spread, you would look up the value in a tangent table. You can see a tangent table with a resolution of 1 degree over <a href="http://www.grc.nasa.gov/WWW/k-12/airplane/tabltan.html">here</a>.
</div>
</div>
<h3>Calculating the Height of a Pyramid and Elevation of the Sun</h3>
<div class="c2">
<div class="c3">
Thales of Miletus measures the length of the shadow cast by the great pyramid and discovers that it is 17 m to the base of the pyramid. Using his 2 m measuring stick he notices that it casts a shadow of 1.8 m. If the great pyramid measures 230 m on a side, how high is the great pyramid? What is the angle of elevation of the sun?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/thales-1-thumb.png" alt="" border="0" /><br />
This is another basic trigonometric problem involving similar triangles, but this time a little trickier to solve. We know the dimensions of the height and length of the <u>measuring-stick-triangle</u>, but we don&#8217;t know the height or length of the <u>pyramid-triangle</u>. We know the length of the shadow to the base of the pyramid, but we need to know the length of the shadow to the center of the pyramid (from which the height is measured).<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/thales-2-thumb.png" alt="" border="0" /><br />
We know the length of a side of the pyramid (this 230 m), we can use this to determine what the length of the shadow to the center of the pyramid should be. The peak of the pyramid is in the center, so it is 115 m from the base of the pyramid. This gives us a corrected shadow length of 17 m to the base + 115 m from the base to the center which is 132 m.<br />
We now know the length of one of the sides of the <u>pyramid-triangle</u>. Since the <u>pyramid-triangle</u> is similar to the <u>measuring-stick-triangle</u>, and we know that similar triangles have the same ratios for their sides, we can calculate the height of the pyramid in the same way as the previous example:</p>
<div class="c1">
In the previous example (Calculating the HEight of a Tree), I wrote the equation as:<br />
<code>a / b = A / B = ratio</code><br />
Equations in mathematics can be written in many different ways. In this example, I don&#8217;t bother to write the ratio and instead write it as:<br />
<code>a / b = A / B</code>
</div>
<p>Substituting the known values we get:<br />
<code>2 / 1.8 = height / 132</code><br />
Rearranging we get:<br />
<code>(2 × 132) / 1.8 = height</code><br />
Simplifying everything we get:<br />
height = 146.7 m<br />
To calculate the elevation of the sun, we know the ratio of height to length. In trigonometric terms, this is the definition of tangent.<br />
tan&alpha; = opposite ÷ adjacent = 2 / 1.8 = 1.111&#8230;<br />
Looking up (or using the inverse key on the calculator) we see that for tan&alpha; to equal 1.111&#8230;, &alpha; must be 48°.
</div>
<h3>Calculating the Length of a Shadow</h3>
<div class="c1">
The Peace Tower in Ottawa, Canada is 92.2 m tall. On July 1st, 2012, at noon, the sun&#8217;s angle of elevation was 67.6°. How long was the shadow cast by the Peace Tower?<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/peace-tower.png" alt="" border="0" /><br />
We know the height of the tower and we know the angle of the sun (relative to the ground).<br />
Unlike the previous two problems, we only have one right angle triangle &#8211; the <u>tower-shadow-triangle</u>. We know if we have two similar triangles, the ratios of their sides must be the same.<br />
We know the angle the sun makes with the ground, so we know one of the angles of the triangle (actually, we know two of them &#8211; the other is 90° &#8211; so, if we need to, we can find the third angle because all angles in a triangle must sum up to 180°).<br />
All right angle triangles with the same angles are similar triangles &#8211; it doesn&#8217;t matter if we don&#8217;t know the lengths of the sides because we know the ratios those sides will have.<br />
We know that the following must be true:<br />
<code>length of the opposite side ÷ length of the adjacent side = tan(&alpha;)</code><br />
Plugging in the values we know, we get:<br />
<code>92.2 / x = tan(67.6°)</code><br />
Evaluating (looking up) <code>tan(67.6°)</code> we get:<br />
<code>9.2. / x = 2.43</code><br />
Rearranging we get<br />
<code>x = 92.2 / 2.43</code><br />
Simplifying we get:<br />
<code>x = 37.94</code><br />
The length of the shadow is 37.94 m (and it is probably safe to round it up to 38 m).
</div>
<h3>Calculating the Height of a Cliff</h3>
<div class="c2">
<div class="c3">
A surveyor measuring the height of the cliff, determines that the angle to the top of the cliff is 60° and the distance to the base of the cliff is 120 m how high is the cliff?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/cliff-60.png" alt="" border="0" /><br />
This is similar to calculating the length of the shadow from two examples earlier.<br />
The distance from the cliff, with the height of the cliff and the straight line to the top of the cliff form a right angle triangle.<br />
We know one of the angles. We also know that all right angle triangles with the same angle are similar triangles &#8211; this means we also know the ratios the various sides of the triangles have.<br />
As before, we know that:<br />
<code>opposite ÷ adjacent = tan(&alpha;)</code><br />
Plugging in the values we know, we get:<br />
<code>x / 120 = tan(60°)</code><br />
Evaluating <code>tan(60°)</code> we get:<br />
<code>x / 120 = 1.732</code><br />
Rearranging we get:<br />
<code>x = 1.732 × 120</code><br />
Simplifying we get:<br />
<code>x = 207.84</code><br />
The cliff is 207.84 m high (it is probably ok to round it to 208 m).</p>
<div class="c3">
How would the problem and solution change if we were given the other angle instead?<br />
In this case, we would have:<br />
<code>120 / x = tan(30°)</code><br />
When solved it would give us the same answer.<br />
The relationship remains the same, the only things that change are which side we call <u>opposite</u> and <u>adjacent</u>.
</div>
</div>
<h3>Calculating the Height of a Lighthouse</h3>
<div class="c1">
<div class="c3">
A surveyor measuring the height of the cliff, determines that the angle to the top of the cliff is 60° &#8211; this angle was measured at a height of 1.7 m (i.e. the surveying equipment is on a tripod and stands 1.7 m above the ground). The angle measured to the top of the lighthouse is 65°. The distance to the cliff is 120 m. How tall is the lighthouse? How high is the cliff?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/cliff-lighthouse.png" alt="" border="0" /><br />
In this case, we are given two triangles, but they are not similar. One triangle is the <u>cliff-triangle</u>, the other is the <u>cliff-lighthouse-triangle</u>.<br />
If we find the height of both triangles, we can subtract the height of the cliff from the height of the cliff + lighthouse and get the height of the lighthouse.</p>
<h3>Cliff Height</h3>
<p>This is basic trigonometric ratios:<br />
<code>cliff height ÷ distance to cliff = tan(60°)</code><br />
Plugging in the known values:<br />
<code>x / 120 = tan(60°)</code><br />
Solving:<br />
<code>x = 207.84</code><br />
We&#8217;re not finished with the height of the cliff, the angle to the top of the cliff was measured from a height of 1.7 m, this means we need to add 1.7 m to the height of the cliff:<br />
<code>height = 207.84 + 1.7 = 209.54 m</code></p>
<div class="c2">
<div class="c3">
If we measured from the base of the tripod, what would the angle be to the top of the cliff?
</div>
<p>In this case, we know the length of two sides of the triangle: 120 m and 209.54 m.<br />
We can calculate the ratio of these two sides:<br />
<code>ratio = 209.54 / 120 = 1.74616...</code><br />
We know from trigonometric definitions that <code>tan(&alpha;) = opposite ÷ adjacent</code>, so all we need to find is what &alpha; gives us the necessary ratio. Using tan<sup>-1</sup> on the calculator we see that:<br />
<code>tan<sup>-1</sup>(1.74616...) = 60.2°</code>
</div>
<h3>Height of the Cliff + Lighthouse</h3>
<p>This is basic trigonometric ratios:<br />
<code>(cliff height + lighthouse height) ÷ distance to cliff = tan(&alpha;)</code><br />
Plugging in the known values we get:<br />
<code>x / 120 = tan(65°)</code><br />
Solving:<br />
<code> x = 257.34</code><br />
We&#8217;re not finished with the height of the cliff + lighthouse, the angle to the top of the cliff + lighthouse was measured from a height of 1.7 m, this means we need to add 1.7 m to the height of the cliff + lighthouse:<br />
<code>x = 257.34 + 1.7 = 259.04 m</code></p>
<h3>Height of the Lighthouse</h3>
<p>It is a simple matter of subtracting the height of the cliff from the height of the cliff + lighthouse:<br />
<code>lighthouse height = (lighthouse height + cliff height) - cliff height</code><br />
Plugging in the known values we get:<br />
<code>x = 259.04 - 209.54</code><br />
Solving we get:<br />
<code>x = 49.5 m</code></p>
<div class="c2">
In this example, it was not necessary to correct the height of the cliff and cliff + lighthouse by adding 1.7 m to each because each was offset (measured) from the same point &#8211; but it is still good practice to find the true heights (dimensions) even if only the relative ones are good enough.
</div>
</div>
<h3>Calculating the Length of a Field #1</h3>
<div class="c2">
<div class="c3">
Two surveyors measure the length of the field. One surveyor stands at one end of the field with the measuring equipment, the other surveyor stands at the other end of the field with a measuring stick 2 m high. The surveyor measures an angle of 2° to the base of the measuring stick. The surveying equipment is at a height of 1.7 m. How long is the field?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-1.png" alt="" border="0" /><br />
This is basic trigonometric ratios. The straight line (of sight) to the measuring stick is perpendicular to the measuring stick, therefore, it forms one side of a right angle triangle. The measuring stick (or part of it) forms another side. Finally, the line of sight to the base of the measuring stick forms the remaining side (which is also the hypotenuse):<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-2.png" alt="" border="0" /><br />
We know the height where the straight line of sight intersects the measuring stick must be 1.7 m. We also know the angle to the base of the measuring stick is 2°.<br />
We don&#8217;t know the length of the side adjacent to the measured angle, we do know the length of the side opposite the measured angle, and we know the angle. Two trigonometric ratios use the <u>opposite</u> and <u>adjacent</u> sides: <em>tangent</em> and <em>cotangent</em>. Since <em>tangent</em> is the more commonly used ratio (and appears on many calculators) we will use the definition of <em>tangent</em>:<br />
<code>tan(&alpha;) = opposite ÷ adjacent</code><br />
Plugging in the known values, we get:<br />
<code>tan(2°) = 1.7 / x</code><br />
Solving, we get:<br />
<code>x = 48.68</code><br />
The length of the field is 48.68 m (it is probably safe to round up to 48.7 m)
</div>
<h3>Calculating the Length of a Field #2</h3>
<div class="c1">
<div class="c3">
Two surveyors measure the length of the field. One surveyor stands at one end of the field with the measuring equipment, the other surveyor stands at the other end of the field with a measuring stick 2 m high. The surveyor forgot to measure the height of the equipment, but, fortunately, took two readings. The surveyor measures an angle of 2° to the base of the measuring stick and 1° to the top of the measuring stick.. How long is the field?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-3.png" alt="" border="0" /><br />
We know the straight line of sight to the measuring stick must be perpendicular to the measuring stick &#8211; this forms one side of a right angle triangle (as in the previous example). In fact, it forms the side of two right angle triangles &#8211; since two different angles were measured. Unlike the previous example, we don&#8217;t know the length of the other side, but we do know the length of the measuring stick &#8211; it is 2 m. We know that the sum of the two opposite sides must be 2 m. We also know the angles for both right angle triangles, so we know the ratios the sides have:<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-4.png" alt="" border="0" /><br />
From trigonometric definitions we know that <code>tan(&alpha;) = opposite / adjacent</code><br />
We also know the <u>adjacent</u> side is the same length for both triangles.<br />
Finally, we know that <code>opposite<sub>top</sub></code> + <code>opposite<sub>bottom</sub></code> = 2 m.</p>
<h3>Top Triangle</h3>
<p>Solving for the top triangle we get:<br />
<code>tan(1°) = opposite<sub>top</sub> / adjacent</code><br />
Calculating what we can, we get:<br />
<code>0.017455 = opposite<sub>top</sub> / adjacent</code></p>
<h3>Bottom Triangle</h3>
<p>Solving for the bottom triangle we get:<br />
<code>tan(2°) = opposite<sub>bottom</sub> / adjacent</code><br />
Calculating what we can, we get:<br />
<code>0.034921 = opposite<sub>top</sub> / adjacent</code></p>
<h3>Combining the Equations</h3>
<div class="c4">
Fundamental to basic algebra is the ability to manipulate an equation any way we like &#8211; as long as we do the same thing on both sides.<br />
For example, consider <span class="i2">4 = 4</span>:<br />
We can add 8 to both sides and the relationship remains the same: <span class="i2">4 + 8 = 4 + 8</span>.<br />
However, it is also perfectly valid to do: <span class="i2">4 + 8 = 4 + 1 + 2 + 5</span> (this is one of those non-intuitive mathematical tricks &#8211; but it is very useful).<br />
We need to employ this algebraic trick of substituting <em>equivalent</em> things into our equations.
</div>
<p>Since we know that the <u>adjacent</u> sides of both right angle triangles are identical, we can use this to merge the two equations together.<br />
Rearranging the bottom triangle equation so the <u>adjacent</u> is alone, we get:<br />
<code>adjacent = opposite<sub>bottom</sub> / 0.034921</code><br />
We now replace the <u>adjacent</u> side in the top triangle with its equivalent which is <code>opposite<sub>bottom</sub> / 0.034921</code>:<br />
<code>0.017455 = opposite<sub>top</sub> / (opposite<sub>bottom</sub> / 0.034921)</code><br />
Rearranging we get:<br />
<code>0.017455 × opposite<sub>bottom</sub> = 0.034921 × opposite<sub>top</sub></code></p>
<h3>Solving for the <em>opposite<sub>bottom</sub></em> side</h3>
<p>Dividing both sides by by 0.017455, we get:<br />
<code>opposite<sub>bottom</sub> = 2 × opposite<sub>top</sub></code></p>
<h3>Remembering the 2 m Measuring Stick</h3>
<p>At this point, we need to remember that <code>opposite<sub>bottom</sub> + opposite<sub>top</sub> = 2</code></p>
<div class="c2">
It is common in mathematics that you have to remember several things at the same time as you proceed through the steps of solving a problem.
</div>
<h3>Algebraic Substitution Trick on the Measuring Stick</h3>
<p>We can do a similar <i>equivalent</i> substitution in this equation (in this case, we will substitute <code>2 × opposite<sub>top</sub></code> for <code>opposite<sub>bottom</sub></code>:<br />
<code>2 × opposite<sub>top</sub> + opposite<sub>top</sub> = 2</code><br />
Solving we get:<br />
<code>opposite<sub>top</sub> = 0.666...</code></p>
<div class="c2">
If the top portion of the stick us 0.666&#8230; m, then the bottom portion must be 1.333&#8230; m
</div>
<h3>Returning to the original equation:</h3>
<div class="c2">
It does not matter which of the lengths we use (as long as we use them in the right place), the answer will still be the same (any small differences in the answers are due to rounding errors &#8211; you can avoid / minimize these by using the full precision provided by your calculator.
</div>
<p>Now we can take this value for <code>opposite<sub>top</sub></code> and substitute it our original equation for the top right angle triangle:<br />
<code>tan (1°) = 0.666... / adjacent</code><br />
Solving we get:<br />
<code>adjacent = 38.19</code><br />
The distance of the field is 38.19 m (which you can, probably, safely round up to 38.2 m)
</div>

<p>The post <a href="https://complete-concrete-concise.com/mathematics/how-to-solve-trigonometric-problems/">How to Solve Trigonometric Problems</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
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