8. Right Triangles & Trigonometry

Lesson

Recall our 3 trigonometric ratios that relate an angle and two sides of a right triangle together.

Trigonometric ratios

$\sin\theta$`s``i``n``θ` = $\frac{Opposite}{Hypotenuse}$`O``p``p``o``s``i``t``e``H``y``p``o``t``e``n``u``s``e`

$\cos\theta$`c``o``s``θ` = $\frac{Adjacent}{Hypotenuse}$`A``d``j``a``c``e``n``t``H``y``p``o``t``e``n``u``s``e`

$\tan\theta$`t``a``n``θ` = $\frac{Opposite}{Adjacent}$`O``p``p``o``s``i``t``e``A``d``j``a``c``e``n``t`

If we know $2$2 side lengths or an angle and a side length, then we can then find any other part of that triangle using these trigonometric ratios.

If we know $2$2 sides and want to find the third, we would use Pythagorean theorem.

Pythagorean theorem

For a right triangle $\triangle ABC$△`A``B``C` with hypotenuse $c$`c`, the following is true about its side lengths:

$a^2+b^2=c^2$`a`2+`b`2=`c`2

If we know $1$1 side length and $1$1 acute angle, we would use one of the trigonometric ratios.

The most common mistake is when the wrong ratio is used. We have to remember the ratios and the sides that apply to those ratios. For most students the mnemonic SOHCAHTOA can be a great help.

In the given triangle $\theta=25^\circ$`θ`=25° and the hypotenuse measures $12.6$12.6. Solve for the length $b$`b` to two decimal places.

**Think**: We need to identify the sides we have and want with respect to the angle given. Here we can see that we have the hypotenuse (H) and we want $b$`b`, which is opposite (O) the angle. This means we have OH - so the trig ratio we need to use here is sine.

**Do**:

$\sin\theta$sinθ |
$=$= | $\frac{O}{H}$OH |
State the formula for the correct ratio |

$\sin25^\circ$sin25° |
$=$= | $\frac{b}{12.6}$b12.6 |
Fill in the given values |

$b$b |
$=$= | $12.6\times\sin25^\circ$12.6×sin25° |
Multiply both sides by 12.6 |

$b$b |
$=$= | $5.32$5.32 |
Simplify |

Find the length of the hypotenuse ($c$`c`) in the diagram, where the angle $36^\circ$36° and the side length of $4.8$4.8 are given, to two decimal places.

**Think**: We need to identify the sides we have and want with respect to the angle given. Here we can see that we want the hypotenuse (H) and we have a side length of $4.8$4.8, which is adjacent (A) the angle. This means we have AH - so the trig ratio we need to use here is cosine.

**Do**:

$\cos\theta$cosθ |
$=$= | $\frac{A}{H}$AH |
State the formula for the correct ratio |

$\cos36^\circ$cos36° |
$=$= | $\frac{4.8}{c}$4.8c |
Fill in the given values |

$c\times\cos36^\circ$c×cos36° |
$=$= | $4.8$4.8 |
Multiply both sides by $c$ |

$c$c |
$=$= | $\frac{4.8}{\cos36^\circ}$4.8cos36° |
Divide both sides by $\cos36^\circ$ |

$c$c |
$=$= | $5.93$5.93 |
Simplify |

Find the length of the unknown side, when the angle is $66^\circ$66° and the indicated side length is $7.3$7.3, to two decimal places.

**Think**: We need to identify the sides we have and want with respect to the angle given. Here we can see that we want the adjacent side ($A$`A`) and we have a side length of $7.3$7.3, which is opposite ($O$`O`) the angle. This means we have $OA$`O``A` - so the trig ratio we need to use here is tangent.

**Do**:

$\tan\theta$tanθ |
$=$= | $\frac{O}{A}$OA |
State the formula for the correct ratio |

$\tan66^\circ$tan66° |
$=$= | $\frac{7.3}{a}$7.3a |
Fill in the given values |

$a$a |
$=$= | $\frac{7.3}{\tan66^\circ}$7.3tan66° |
Cross multiply (really multiplying by $a$ |

$a$a |
$=$= | $3.25$3.25 |
Simplify |

The vast majority of the time, we will be asked to round to a specified number of decimal places. However, if the acute angle is $30^\circ$30°, $45^\circ$45° or $60^\circ$60°, we can use the exact values. Our answer may involve a radical.

Trigonometric ratios and special angles

$30^\circ$30° | $45^\circ$45° | $60^\circ$60° | |
---|---|---|---|

sin | $\sin30^\circ=\frac{1}{2}$sin30°=12 |
$\sin45^\circ=\frac{\sqrt{2}}{2}$sin45°=√22 |
$\sin60^\circ=\frac{\sqrt{3}}{2}$sin60°=√32 |

cos | $\cos30^\circ=\frac{\sqrt{3}}{2}$cos30°=√32 |
$\cos45^\circ=\frac{\sqrt{2}}{2}$cos45°=√22 |
$\cos60^\circ=\frac{1}{2}$cos60°=12 |

tan | $\tan30^\circ=\frac{\sqrt{3}}{3}$tan30°=√33 |
$\tan45^\circ=1$tan45°=1 |
$\tan60^\circ=\sqrt{3}$tan60°=√3 |

Find the value of $f$`f`, correct to two decimal places.

Find the value of $x$`x`, the side length of the parallelogram, to the nearest centimeter.

Find the length of side $h$`h`.

Use sine, cosine, tangent, the Pythagorean Theorem and properties of special right triangles to solve right triangles in applied problems. ★