Hong Kong
Stage 4 - Stage 5

# Applications to Real Life

Lesson

We are starting to see a wide variety of trigonometric problems that required us to solve for unknown lengths (distances, heights, and so on) and angles (including angles of elevation and depression).

The following set of questions uses all of our trigonometric knowledge thus far, this summary may be of use.

Right-angled Triangles

Pythagoras' theorem:  $a^2+b^2=c^2$a2+b2=c2, where $c$c is the hypotenuse

$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse = $\frac{O}{H}$OH

$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse = $\frac{A}{H}$AH

$\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent =$\frac{O}{A}$OA

Angle of Elevation:  the angle made between the line of sight of the observer and the 'horizontal' when the object is ABOVE the horizontal (observer is looking UP

Angle of Depression:  the angle made between the line of sight of the observer and the 'horizontal' when the object is BELOW the horizontal (observer is looking DOWN)

Exact value triangles

Here is a worked solution to a practical application.

#### Example

A ship is $27$27m away from the bottom of a cliff. A lighthouse is located at the top of the cliff. The ship's distance is $34$34m from the bottom of the lighthouse and $37$37m from the top of the lighthouse.

1. Find the distance from the bottom of the cliff to the top of the lighthouse, $y$y, correct to two decimal places.

2. Find the distance from the bottom of the cliff to the bottom of the lighthouse, $x$x, correct to two decimal places.

3. Hence find the height of the lighthouse to the nearest tenth of a metre.