AustraliaNSW
Stage 5.1-2

# 8.05 Applications of trigonometry

Lesson

The trigonometric ratios gave us relationships between the sides and angles in a right-angled triangle. In real life situations we won't always know all the information about a right-angled triangle and we can use the trigonometric ratios to find those missing values.

### Finding angles using trigonometry

To find missing angles in right-angled triangles using trigonometry, we require at least two known sides and the trigonometric ratios.

Remember that the trigonometric ratios for a right-angled triangle are:

• $\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse
• $\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse
• $\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent

To isolate the angle in each of these relationships, we can apply the inverse trigonometric function to each side of the equation. This will give us:

• $\theta=\sin^{-1}\left(\frac{\text{Opposite }}{\text{Hypotenuse }}\right)$θ=sin1(Opposite Hypotenuse )
• $\theta=\cos^{-1}\left(\frac{\text{Adjacent }}{\text{Hypotenuse }}\right)$θ=cos1(Adjacent Hypotenuse )
• $\theta=\tan^{-1}\left(\frac{\text{Opposite }}{\text{Adjacent }}\right)$θ=tan1(Opposite Adjacent )

Any of these relationships can be used to find $\theta$θ depending on which side lengths of the triangle are known.

Caution

While we may represent the inverse trigonometric functions using an index of $-1$1, they are not reciprocals of the original functions.

For example: $\sin^{-1}\theta$sin1θ $\ne$ $\frac{1}{\sin\theta}$1sinθ

### Finding side lengths using trigonometry

To find missing sides in right-angled triangles using trigonometry, we require at least one angle (other than the right angle) and at least one other side length.

We can find a missing side length using one of the trigonometric ratios where the missing length the only unknown value. Doing this allows us to solve the equation and find that value.

For example:

In this right-angled triangle, we are given one angle and one side. Using this, we want to find the length $x$x.

Since the given side is adjacent to the given angle and $x$x is the length of the hypotenuse, we can express their relationship using the trigonometric ratio:

$\cos26^\circ=\frac{5}{x}$cos26°=5x

We can isolate $x$x in this equation by multiplying both sides by $x$x and then dividing both sides by $\cos26^\circ$cos26°. This will give us:

$x=\frac{5}{\cos26^\circ}$x=5cos26°

Solving this tells us that $x=5.56$x=5.56, rounded to two decimal places.

Depending on the given angle, given side and missing side, we will need to use one of the three trigonometric ratios so that all relevant information is in the equation.

#### Practice questions

##### Question 1

The person in the picture sights a pigeon above him. If the angle the person is looking at is $\theta$θ, find $\theta$θ in degrees.

##### Question 2

A helicopter is $344$344 metres away from its landing pad.

If the angle of depression to the landing pad is $32^\circ$32°, what is the height, $x$x, of the helicopter above the ground?

##### Question 3

A $27.3$27.3 m long string of lights joins the top of a tree to a point on the ground.

If the tree is $7.4$7.4 m tall, find $\theta$θ, the angle the string of lights would make with the tree to the nearest two decimal places.

### Outcomes

#### MA5.1-10MG

applies trigonometry, given diagrams, to solve problems, including problems involving angles of elevation and depression