randRange( 2, 7 )

In the right triangle shown, AC = BC = AC. What is AB?

betterTriangle( 1, 1, "A", "B", "C", AC, AC, "x" );
AC * AC * 2

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?

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.

Let's try using sine:
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");
Sine is opposite over hypotenuse (SOH CAH TOA), so \sin {45}^{\circ} must be \dfrac{AC}{x}. We also know that \sin{45}^{\circ} = \dfrac{\sqrt{2}}{2}.
Solving for x, we get

\qquad x \cdot \sin {45}^{\circ} = AC

\qquad x \cdot \dfrac{\sqrt{2}}{2} = AC

\qquad x = AC \cdot \dfrac{2}{\sqrt{2}}

So, we see that the hypotenuse is \sqrt{2} times as long as each of the legs, since x = AC \cdot \sqrt{2}.

2 * randRange( 2, 6 )

In the right triangle shown, AC = BC and AB = AB. How long are each of the legs?

betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB );
AB * AB / 2

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?

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.

Let's try using cosine:
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");
Cosine is adjacent over hypotenuse (SOH CAH TOA), so \cos {45}^{\circ} must be \dfrac{x}{AB}. We also know that \cos{45}^{\circ} = \dfrac{\sqrt{2}}{2}.

Solving for x, we get x = AB \cdot \cos {45}^{\circ} = AB \cdot \dfrac{\sqrt{2}}{2}

So, x = AB/2 \sqrt{2}.

2 * randRange( 2, 6 )

In the right triangle shown, AC = BC and AB = AB\sqrt{2}. How long are each of the legs?

betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB + "\\sqrt{2}" );
AB * AB

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?

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.

Let's try using cosine:
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");
Cosine is adjacent over hypotenuse (SOH CAH TOA), so \cos {45}^{\circ} must be \dfrac{x}{AB\sqrt{2}}. We also know that \cos{45}^{\circ} = \dfrac{\sqrt{2}}{2}.

Solving for x, we get x = AB\sqrt{2} \cdot \cos {45}^{\circ} = AB\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}

So, x = AB \left(\dfrac{\sqrt{2}\cdot\sqrt{2}}{2}\right) = AB \left(\dfrac{2}{2}\right) = AB.

randRange( 2, 6 ) randFromArray([ [1, ""], [3, "\\sqrt{3}"] ]) BC + BCrs randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])

In the right triangle shown, mAB and BC = BC + BCrs. How long is AB?

betterTriangle( 1, sqrt(3), "A", "B", "C", BC + BCrs, "", "x" );
4 * BC * BC * BCr

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?

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.

Let's try using sine:
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");
Sine is opposite over hypotenuse (SOH CAH TOA), so \sin {30}^{\circ} = \dfrac{BCdisp}{x}. We also know that \sin{30}^{\circ} = \dfrac{1}{2}.
Solving for x, we get

\qquad x \cdot \sin{30}^{\circ} = BCdisp

\qquad x \cdot \dfrac{1}{2} = BCdisp

\qquad x = BCdisp \cdot 2

So, x = BC*2 + BCrs.

3 * randRange( 2, 6 ) randFromArray([ [1, "", (AC * 2 / 3) + "\\sqrt{3}", AC * AC * 4 / 3], [3, "\\sqrt{3}", (AC * 2), AC * AC * 4] ]) AC + ACrs randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])

In the right triangle shown, mAB and AC = AC + ACrs. How long is AB?

betterTriangle( 1, sqrt(3), "A", "B", "C", "", AC + ACrs, "x" );
AB

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?

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.

Let's try using cosine:
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");
Cosine is adjacent over hypotenuse (SOH CAH TOA), so \cos {30}^{\circ} = \dfrac{ACdisp}{x}. We also know that \cos{30}^{\circ} = \dfrac{\sqrt{3}}{2}.
Solving for x, we get

\qquad x \cdot \cos{30}^{\circ} = ACdisp

\qquad x \cdot \dfrac{\sqrt{3}}{2} = ACdisp

\qquad x = ACdisp \cdot \dfrac{2}{\sqrt{3}}

\qquad x = ACdisp \cdot \dfrac{2\cdot\sqrt{3}}{3}

So, x = ABs.

randRange( 2, 6 ) randFromArray([ [1, ""], [3, "\\sqrt{3}"] ]) 2*BC + BCrs randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])

In the right triangle shown, mAB and AB = ( 2 * BC ) + BCrs. How long is BC?

betterTriangle( 1, sqrt(3), "A", "B", "C", "x", "", ( 2 * BC ) + BCrs );
BC * BC * BCr

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?

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.

Let's try using cosine:
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");
Cosine is adjacent over hypotenuse (SOH CAH TOA), so \cos {60}^{\circ} = \dfrac{x}{ABdisp}. We also know that \cos{60}^{\circ} = \dfrac{1}{2}.
Solving for x, we get

\qquad x = ABdisp \cdot \cos{60}^{\circ}

\qquad x = ABdisp \cdot \dfrac{1}{2}

So, x = BC + BCrs.

3 * randRange( 2, 6 ) randFromArray([ [1, "", (AC * 2 / 3) + "\\sqrt{3}", AC * AC * 4 / 3], [3, "\\sqrt{3}", (AC * 2), AC * AC * 4] ]) randFromArray([ "\\angle A = 30^\\circ", "\\angle B = 60^\\circ" ])

In the right triangle shown, mAB and AB = ABs. How long is AC?

betterTriangle( 1, sqrt(3), "A", "B", "C", "", "x", ABs );
AC * AC * ACr

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?

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.

Let's try using sine:
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");
Sine is opposite over hypotenuse (SOH CAH TOA), so \sin {60}^{\circ} = \dfrac{x}{ABs}. We also know that \sin{60}^{\circ} = \dfrac{\sqrt{3}}{2}.
Solving for x, we get

\qquad x = ABs \cdot \sin{60}^{\circ}

\qquad x = ABs \cdot \dfrac{\sqrt{3}}{2}

So, x = AC + ACrs.