In the right triangle shown, `AC = BC = `

. What is `AC``AB?`

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

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}

In the right triangle shown, `AC = BC`

and `AB = `

. How long are each of the legs?`AB`

betterTriangle( 1, 1, "A", "B", "C", "x", "x", 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 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}

In the right triangle shown, `AC = BC`

and `AB = `

. How long are each of the legs?`AB`\sqrt{2}

betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB + "\\sqrt{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 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`

In the right triangle shown,

and `mAB``BC = `

. How long is `BC + BCrs``AB?`

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

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`

In the right triangle shown,

and `mAB``AC = `

. How long is `AC + ACrs``AB?`

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

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`

In the right triangle shown,

and `mAB``AB = `

. How long is `( 2 * BC ) + BCrs``BC?`

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

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`

In the right triangle shown,

and `mAB``AB = `

. How long is `ABs``AC?`

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

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?

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`