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pythagorean_theorem_2.html
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pythagorean_theorem_2.html
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<!DOCTYPE html>
<html data-require="math graphie math-format">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Special right triangles</title>
<script src="../khan-exercise.js"></script>
<script>
function betterTriangle(width, height, A, B, C, a, b, c) {
var scale = 5 / Math.sqrt(width * width + height * height);
width *= scale;
height *= scale;
with ( KhanUtil.currentGraph ) {
// Leave some space for the labels
init({ range: [[-1.5, width + 1], [-1, height + 1]] });
path([ [0, 0], [width, 0], [0, height], true ]);
label( [0, height], A, "above left" );
label( [0, 0], C, "below left" );
label( [width, 0], B, "below right" );
label( [0, height/2], b, "left" );
label( [width/2, 0], a, "below" );
label( [width/2, height/2], c, "above right", {
labelDistance: 3
} );
}
}
</script>
</head>
<body>
<div class="exercise">
<div class="problems">
<div id="45-45-90-find-hypotenuse">
<div class="vars">
<var id="AC">randRange( 2, 7 )</var>
</div>
<div class="question">
<p>In the right triangle shown, <code>AC = BC = <var>AC</var></code>. What is <code>AB?</code></p>
<div class="graphie">
betterTriangle( 1, 1, "A", "B", "C", AC, AC, "x" );
</div>
</div>
<div class="solution" data-type="radical"><var>AC * AC * 2</var></div>
<div class="hints">
<p>
We know the length of each leg, and want to
find the length of the hypotenuse. What
mathematical relationship
is there between a right triangle's leg and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). Since the two
legs of this triangle are congruent, this is a 45-45-90
triangle, and we know what the values of sine and cosine
are at each angle of the triangle.
</p>
<div>
Let's try using sine:
<div class="graphie">
betterTriangle( 1, 1, "A", "B", "C", AC, AC, "x" );
arc([5/sqrt(2), 0], 0.5, 135, 180);
label([5/sqrt(2)-0.4, -0.1],
"{45}^{\\circ}", "above left");
</div>
Sine is opposite over hypotenuse (SOH CAH TOA), so
<code>\sin {45}^{\circ}</code> must be
<code>\dfrac{<var>AC</var>}{x}</code>. We also know that
<code>\sin{45}^{\circ} = \dfrac{\sqrt{2}}{2}</code>.
</div>
<div>
Solving for <code>x</code>, we get
<p><code>\qquad x \cdot \sin {45}^{\circ} =
<var>AC</var></code></p>
<p><code>\qquad x \cdot \dfrac{\sqrt{2}}{2} =
<var>AC</var></code></p>
<p><code>\qquad x = <var>AC</var> \cdot
\dfrac{2}{\sqrt{2}}</code></p>
</div>
<p>
So, we see that the hypotenuse is <code>\sqrt{2}</code> times as long as each of the legs, since
<code>x = <var>AC</var> \cdot \sqrt{2}</code>.
</p>
</div>
</div>
<div id="45-45-90-find-leg" data-weight="0.5">
<div class="vars">
<var id="AB">2 * randRange( 2, 6 )</var>
</div>
<div class="question">
<p>In the right triangle shown, <code>AC = BC</code> and <code>AB = <var>AB</var></code>. How long are each of the legs?</p>
<div class="graphie">
betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB );
</div>
</div>
<div class="solution" data-type="radical"><var>AB * AB / 2</var></div>
<div class="hints">
<p>
We know the length of the hypotenuse, and want to
find the length of each leg.
What mathematical relationship
is there between a right triangle's legs and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). Since the two
legs of this triangle are congruent, this is a 45-45-90
triangle, and we know what the values of sine and cosine
are at each angle of the triangle.
</p>
<div>
Let's try using cosine:
<div class="graphie">
betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB );
arc([5/sqrt(2), 0], 0.5, 135, 180);
label([5/sqrt(2)-0.4, -0.1],
"{45}^{\\circ}", "above left");
</div>
Cosine is adjacent over hypotenuse (SOH CAH TOA), so
<code>\cos {45}^{\circ}</code> must be
<code>\dfrac{x}{<var>AB</var>}</code>. We also know that
<code>\cos{45}^{\circ} = \dfrac{\sqrt{2}}{2}</code>.
</div>
<p>
Solving for <code>x</code>, we get
<code>x = <var>AB</var> \cdot \cos {45}^{\circ}
= <var>AB</var> \cdot \dfrac{\sqrt{2}}{2}</code>
</br>
</p>
<p>
So, <code>x = <var>AB/2</var> \sqrt{2}</code>.
</p>
</div>
</div>
<div id="45-45-90-find-leg-2" data-weight="0.5">
<div class="vars">
<var id="AB">2 * randRange( 2, 6 )</var>
</div>
<div class="question">
<p>In the right triangle shown, <code>AC = BC</code> and <code>AB = <var>AB</var>\sqrt{2}</code>. How long are each of the legs?</p>
<div class="graphie">
betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB + "\\sqrt{2}" );
</div>
</div>
<div class="solution" data-type="radical"><var>AB * AB</var></div>
<div class="hints">
<p>
We know the length of the hypotenuse, and want to
find the length of each leg.
What mathematical relationship
is there between a right triangle's legs and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). Since the two
legs of this triangle are congruent, this is a 45-45-90
triangle, and we know the value of sine and cosine at each
angle of the triangle.
</p>
<div>
Let's try using cosine:
<div class="graphie">
betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB +
"\\sqrt{2}" );
arc([5/sqrt(2), 0], 0.5, 135, 180);
label([5/sqrt(2)-0.4, -0.1],
"{45}^{\\circ}", "above left");
</div>
Cosine is adjacent over hypotenuse (SOH CAH TOA), so
<code>\cos {45}^{\circ}</code> must be
<code>\dfrac{x}{<var>AB</var>\sqrt{2}}</code>. We also know
that <code>\cos{45}^{\circ} = \dfrac{\sqrt{2}}{2}</code>.
</div>
<p>
Solving for <code>x</code>, we get
<code>x = <var>AB</var>\sqrt{2} \cdot \cos {45}^{\circ}
= <var>AB</var>\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}</code>
</br>
</p>
<p>
So, <code>x = <var>AB</var>
\left(\dfrac{\sqrt{2}\cdot\sqrt{2}}{2}\right)
= <var>AB</var>
\left(\dfrac{2}{2}\right) = <var>AB</var></code>.
</p>
</div>
</div>
<div id="30-60-90-find-hypotenuse-given-short-leg">
<div class="vars">
<var id="BC">randRange( 2, 6 )</var>
<var id="BCr, BCrs">randFromArray([ [1, ""], [3, "\\sqrt{3}"] ])</var>
<var id="BCdisp">BC + BCrs</var>
<var id="mAB">randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])</var>
</div>
<div class="question">
<p>In the right triangle shown, <code><var>mAB</var></code> and <code>BC = <var>BC + BCrs</var></code>. How long is <code>AB?</code></p>
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", BC + BCrs, "", "x" );
</div>
</div>
<div class="solution" data-type="radical"><var>4 * BC * BC * BCr</var></div>
<div class="hints">
<p>
We know the length of a leg, and want to find the length
of the hypotenuse.
What mathematical relationship
is there between a right triangle's legs and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). This is a
30-60-90 triangle, so we know what the values of sine
and cosine are at each angle of the triangle.
</p>
<div>
Let's try using sine:
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", BC + BCrs, "", "x" );
arc([0, 5*sqrt(3)/2], 0.8, 270, 300);
label([-0.1, (5*sqrt(3)/2)-1],
"{30}^{\\circ}", "below right");
</div>
Sine is opposite over hypotenuse (SOH CAH TOA), so
<code>\sin {30}^{\circ} =
\dfrac{<var>BCdisp</var>}{x}</code>. We also know
that <code>\sin{30}^{\circ} = \dfrac{1}{2}</code>.
</div>
<div>
Solving for <code>x</code>, we get
<p><code>\qquad x \cdot \sin{30}^{\circ} =
<var>BCdisp</var></code></p>
<p><code>\qquad x \cdot \dfrac{1}{2} =
<var>BCdisp</var></code></p>
<p><code>\qquad x = <var>BCdisp</var>
\cdot 2</code></p>
</div>
<p>
So, <code>x = <var>BC*2 + BCrs</var></code>.
</p>
</div>
</div>
<div id="30-60-90-find-hypotenuse-given-long-leg">
<div class="vars">
<var id="AC">3 * randRange( 2, 6 )</var>
<var id="ACr, ACrs, ABs, AB">randFromArray([
[1, "", (AC * 2 / 3) + "\\sqrt{3}", AC * AC * 4 / 3],
[3, "\\sqrt{3}", (AC * 2), AC * AC * 4]
])</var>
<var id="ACdisp">AC + ACrs</var>
<var id="mAB">randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])</var>
</div>
<div class="question">
<p>In the right triangle shown, <code><var>mAB</var></code> and <code>AC = <var>AC + ACrs</var></code>. How long is <code>AB?</code></p>
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", "", AC + ACrs, "x" );
</div>
</div>
<div class="solution" data-type="radical"><var>AB</var></div>
<div class="hints">
<p>
We know the length of a leg, and want to find the length
of the hypotenuse.
What mathematical relationship
is there between a right triangle's legs and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). This is a
30-60-90 triangle, so we know what the values of sine
and cosine are at each angle of the triangle.
</p>
<div>
Let's try using cosine:
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", "", AC + ACrs, "x" );
arc([0, 5*sqrt(3)/2], 0.8, 270, 300);
label([-0.1, (5*sqrt(3)/2)-1],
"{30}^{\\circ}", "below right");
</div>
Cosine is adjacent over hypotenuse (SOH CAH TOA), so
<code>\cos {30}^{\circ} =
\dfrac{<var>ACdisp</var>}{x}</code>. We also know
that <code>\cos{30}^{\circ} = \dfrac{\sqrt{3}}{2}</code>.
</div>
<div>
Solving for <code>x</code>, we get
<p><code>\qquad x \cdot \cos{30}^{\circ} =
<var>ACdisp</var></code></p>
<p><code>\qquad x \cdot \dfrac{\sqrt{3}}{2} =
<var>ACdisp</var></code></p>
<p><code>\qquad x = <var>ACdisp</var> \cdot
\dfrac{2}{\sqrt{3}}</code></p>
<p><code>\qquad x = <var>ACdisp</var> \cdot
\dfrac{2\cdot\sqrt{3}}{3}</code></p>
</div>
<p>
So, <code>x = <var>ABs</var></code>.
</p>
</div>
</div>
<div id="30-60-90-find-short-leg-given-hypotenuse">
<div class="vars">
<var id="BC">randRange( 2, 6 )</var>
<var id="BCr, BCrs">randFromArray([ [1, ""], [3, "\\sqrt{3}"] ])</var>
<var id="ABdisp">2*BC + BCrs</var>
<var id="mAB">randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])</var>
</div>
<div class="question">
<p>In the right triangle shown, <code><var>mAB</var></code> and <code>AB = <var>( 2 * BC ) + BCrs</var></code>. How long is <code>BC?</code></p>
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", "x", "", ( 2 * BC ) + BCrs );
</div>
</div>
<div class="solution" data-type="radical"><var>BC * BC * BCr</var></div>
<div class="hints">
<p>
We know the length of the hypotenuse of this triangle,
and want to find the length of a leg.
What mathematical relationship
is there between a right triangle's legs and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). This is a
30-60-90 triangle, so we know what the values of sine
and cosine are at each angle of the triangle.
</p>
<div>
Let's try using cosine:
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", "x", "", ( 2 * BC ) + BCrs );
arc([5/2,0], 0.5, 120, 180);
label([5/2-0.2, 0],
"{60}^{\\circ}", "above left");
</div>
Cosine is adjacent over hypotenuse (SOH CAH TOA), so
<code>\cos {60}^{\circ} =
\dfrac{x}{<var>ABdisp</var>}</code>. We also know
that <code>\cos{60}^{\circ} = \dfrac{1}{2}</code>.
</div>
<div>
Solving for <code>x</code>, we get
<p><code>\qquad x = <var>ABdisp</var>
\cdot \cos{60}^{\circ}</code></p>
<p><code>\qquad x = <var>ABdisp</var> \cdot
\dfrac{1}{2}</code></p>
</div>
<p>
So, <code>x = <var>BC + BCrs</var></code>.
</p>
</div>
</div>
<div id="30-60-90-find-long-leg-given-hypotenuse">
<div class="vars">
<var id="AC">3 * randRange( 2, 6 )</var>
<var id="ACr, ACrs, ABs, AB">randFromArray([
[1, "", (AC * 2 / 3) + "\\sqrt{3}", AC * AC * 4 / 3],
[3, "\\sqrt{3}", (AC * 2), AC * AC * 4]
])</var>
<var id="mAB">randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])</var>
</div>
<div class="question">
<p>In the right triangle shown, <code><var>mAB</var></code> and <code>AB = <var>ABs</var></code>. How long is <code>AC?</code></p>
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", "", "x", ABs );
</div>
</div>
<div class="solution" data-type="radical"><var>AC * AC * ACr</var></div>
<div class="hints">
<p>
We know the length of the hypotenuse of this triangle,
and want to find the length of a leg.
What mathematical relationship
is there between a right triangle's legs and its
hypotenuse?
</p>
<p>
We can use either sine (opposite leg over hypotenuse)
or cosine (adjacent leg over hypotenuse). This is a
30-60-90 triangle, so we know what the values of sine
and cosine are at each angle of the triangle.
</p>
<div>
Let's try using sine:
<div class="graphie">
betterTriangle( 1, sqrt(3), "A", "B", "C", "", "x", ABs );
arc([5/2,0], 0.5, 120, 180);
label([5/2-0.2, 0],
"{60}^{\\circ}", "above left");
</div>
Sine is opposite over hypotenuse (SOH CAH TOA), so
<code>\sin {60}^{\circ} =
\dfrac{x}{<var>ABs</var>}</code>. We also know
that <code>\sin{60}^{\circ} = \dfrac{\sqrt{3}}{2}</code>.
</div>
<div>
Solving for <code>x</code>, we get
<p><code>\qquad x = <var>ABs</var>
\cdot \sin{60}^{\circ}</code></p>
<p><code>\qquad x = <var>ABs</var> \cdot
\dfrac{\sqrt{3}}{2}</code></p>
</div>
<p>
So, <code>x = <var>AC + ACrs</var></code>.
</p>
</div>
</div>
</div>
</div>
</body>
</html>