Trigonometry

UK Secondary (7-11)

Applications to Geometry

Lesson

So far we have found unknown side lengths using Pythagoras' theorem and then looked at 3 special ratios that we can use to find unknown sides or angles in right-angled triangles.

Right-angled triangles

Pythagoras' theorem: $a^2+b^2=c^2$`a`2+`b`2=`c`2, where $c$`c` is the hypotenuse

$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$`s``i``n``θ`=Opposite Hypotenuse = $\frac{O}{H}$`O``H`

$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$`c``o``s``θ`=Adjacent Hypotenuse = $\frac{A}{H}$`A``H`

$\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$`t``a``n``θ`=Opposite Adjacent =$\frac{O}{A}$`O``A`

Problem solving in trigonometry can be in finding unknowns like we have already been doing, using trigonometry in real world applications or in solving geometrical problems like these.

Find $x$`x` in the following geometrical diagram,

**Think**: In order to find $x$`x`, I will need to identify some other measurements along the way. My problem solving strategy will be

1. Find length $AC$`A``C` using trig ratio sine

2. Find length $ED$`E``D`, $\frac{AC}{3}$`A``C`3

3. Find length $x$`x`, using trig ratio sine

**Do**:

1. Find length $AC$`A``C` using trig ratio sine

$\sin23^\circ$sin23° |
$=$= | $\frac{43.6}{AC}$43.6AC |

$AC$AC |
$=$= | $\frac{43.6}{\sin23^\circ}$43.6sin23° |

$AC$AC |
$=$= | $111.59$111.59 |

2. Find length $ED$`E``D`, $\frac{AC}{3}$`A``C`3

$ED=\frac{111.59}{3}$`E``D`=111.593

$ED=37.2$`E``D`=37.2

3. Find length $x$`x`, using trig ratio sine

$\sin35.6^\circ$sin35.6° |
$=$= | $\frac{x}{37.2}$x37.2 |

$x$x |
$=$= | $37.2\times\sin35.6^\circ$37.2×sin35.6° |

$x$x |
$=$= | $21.65$21.65 |

Consider the following diagram.

What is the value of $x$

`x`? Give your answer correct to 2 decimal places.Using the rounded value of $x$

`x`, find the value of $y$`y`. Give your answer correct to 2 decimal places.

Find the length of the unknown side, x, in the given trapezium.

Give your answer correct to $2$2 decimal places.