Trigonometry

Hong Kong

Stage 1 - Stage 3

Lesson

As we saw in our chapter on Special Ratios, there are 3 trigonometric ratios that relate an angle and sides together. They are:

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 any $2$2 parts of a right-angled triangle, whether that's $2$2 side lengths, or an angle and a side length 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 formula

Pythagoras

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

If we know 1 side length and an 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`.

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

**Do**:

$\sin\theta$sinθ |
$=$= | $\frac{O}{H}$OH |

$\sin25^\circ$sin25° |
$=$= | $\frac{b}{12.6}$b12.6 |

$b$b |
$=$= | $12.6\times\sin25^\circ$12.6×sin25° |

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

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.

**Think**: We need to identify the sides we have and want with respect to the angle given. Here I 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 I have AH - so the trig ratio I need to use here is cosine.

**Do**:

$\cos\theta$cosθ |
$=$= | $\frac{A}{H}$AH |

$\cos36^\circ$cos36° |
$=$= | $\frac{4.8}{c}$4.8c |

$c$c |
$=$= | $\frac{4.8}{\cos36^\circ}$4.8cos36° |

$c$c |
$=$= | $5.93$5.93 units (to 2 d.p.) |

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

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

**Do**:

$\tan\theta$tanθ |
$=$= | $\frac{O}{A}$OA |

$\tan66^\circ$tan66° |
$=$= | $\frac{7.3}{a}$7.3a |

$a$a |
$=$= | $\frac{7.3}{\tan66^\circ}$7.3tan66° |

$a$a |
$=$= | $3.25$3.25 units (to 2 d.p.) |

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

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

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