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Actual Hints for Pythagorean Theorem 2

Reviewers: eater

Reviewed By: eater

CC: eater, emily

Differential Revision: http://phabricator.khanacademy.org/D567
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commit c069833ee91cf5cbf0e6acf01d92428f0ae8efd3 1 parent e66dea6
@mwittels mwittels authored
Showing with 282 additions and 37 deletions.
  1. +282 −37 exercises/pythagorean_theorem_2.html
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319 exercises/pythagorean_theorem_2.html
@@ -2,7 +2,7 @@
<html data-require="math graphie math-format">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
- <title>Pythagorean theorem 2</title>
+ <title>Special right triangles</title>
<script src="../khan-exercise.js"></script>
<script>
function betterTriangle(width, height, A, B, C, a, b, c) {
@@ -38,19 +38,56 @@
</div>
<div class="question">
- <p>In the right triangle shown, <code>AC = BC = <var>AC</var></code>. What is <code>AB</code>?</p>
+ <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, "?" );
+ 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>This is a 45-45-90 triangle because the two legs are congruent.</p>
- <p>Thus, the hypotenuse is <code>\sqrt{2}</code> times as long as each of the legs.</p>
- <p><code>AB = <var>AC</var>\sqrt{2}</code></p>
+ <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>
@@ -63,16 +100,49 @@
<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", "?", "?", AB );
+ 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>This is a 45-45-90 triangle because the two legs are congruent.</p>
- <p>Thus, each leg is <code>\frac{1}{\sqrt{2}}</code> times as long as the hypotenuse.</p>
- <p><code>AC = BC = \frac{1}{\sqrt{2}} <var>AB</var> = <var>AB / 2</var> \sqrt{2}</code></p>
+ <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>
@@ -85,16 +155,53 @@
<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", "?", "?", AB + "\\sqrt{2}" );
+ 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>This is a 45-45-90 triangle because the two legs are congruent.</p>
- <p>Thus, each leg is <code>\frac{1}{\sqrt{2}}</code> times as long as the hypotenuse.</p>
- <p><code>AC = BC = \frac{1}{\sqrt{2}} <var>AB</var> \sqrt{2} = <var>AB</var></code></p>
+ <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>
@@ -102,24 +209,59 @@
<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>
+ <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, "", "?" );
+ 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>This is a 30-60-90 triangle.</p>
- <p>Thus, the hypotenuse is twice as long as the shorter leg.</p>
- <p><code>AB = 2 \cdot BC = 2 \cdot (<var>BC + BCrs</var>) = <var>( 2 * BC ) + BCrs</var></code></p>
+ <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>
@@ -130,24 +272,61 @@
[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>
+ <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, "?" );
+ 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>This is a 30-60-90 triangle.</p>
- <p>Thus, the hypotenuse is <code>\frac{2}{\sqrt{3}}</code> times as long as the longer leg.</p>
- <p><code>AB = \frac{2}{\sqrt{3}} \cdot AC = \frac{2}{\sqrt{3}} \cdot (<var>AC + ACrs</var>) = <var>ABs</var></code></p>
+ <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>
@@ -155,24 +334,57 @@
<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>
+ <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", "?", "", ( 2 * BC ) + BCrs );
+ 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>This is a 30-60-90 triangle.</p>
- <p>Thus, the shorter leg is half as long as the hypotenuse.</p>
- <p><code>BC = \frac12 \cdot AB = \frac12 \cdot (<var>( 2 * BC ) + BCrs</var>) = <var>( BC ) + BCrs</var></code></p>
+ <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>
@@ -188,19 +400,52 @@
</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>
+ <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", "", "?", ABs );
+ 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>This is a 30-60-90 triangle.</p>
- <p>Thus, the longer leg is <code>\frac{\sqrt{3}}{2}</code> times as long as the hypotenuse.</p>
- <p><code>AC = \frac{\sqrt{3}}{2} \cdot AB = \frac{\sqrt{3}}{2} \cdot (<var>ABs</var>) = <var>AC + ACrs</var></code></p>
+ <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>
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