 AustraliaNSW
Stage 5.1-2

# 8.09 Problem solving with trigonometry

Lesson

In real life situations involving right-angled triangles, we can use trigonometry to solve problem we normally couldn't by finding angles and lengths that are otherwise impossible to calculate.

### 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 is 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 by expressing it in a trigonometric ratio as the only unknown value, and then solve the equation to 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.

### Multiple applications of trigonometry

In some cases, a single application of trigonometry on a single right-angled triangle will not be enough to find the values we are looking for.

In these cases, we can use trigonometry once to find a new value, and then use trigonometry again with our new value to find another new value. We can repeat this process as many times as is necessary to find the value that we are looking for.

#### Exploration

Karen is standing $30$30 m from the base of a tower with an angle of elevation of $27^\circ$27° from the top of the tower. If Nick's angle of elevation is $44^\circ$44° from the top of the tower, what is his distance $x$x from the base of the tower? In order to work out the distance from Nick to the base of the tower, we need to know either the height of the tower or Nick's distance from the top of the tower.

Since we are given the distance from Karen to the base of the tower, we can apply trigonometry once to find the height of the tower, then this value to find our desired value.

Using trigonometry once, we can calculate the height $h$h of the tower to be:

$h=30\tan27^\circ$h=30tan27°

Using this new value, we can find the distance from Nick to the base of the tower $x$x (rounding to two decimal places) to be:

$x$x $=$= $\frac{h}{\tan44^\circ}$htan44° $=$= $\frac{30\tan27^\circ}{\tan44^\circ}$30tan27°tan44° $=$= $15.83$15.83

### Simultaneous applications of trigonometry

In other cases, multiple applications of trigonometry need to be used at the same time in order to solve for a common factor.

These cases arise when multiple trigonometric ratios involve the same missing value. If some relationship between these ratios is known then the missing value can be solved for.

#### Exploration

Joshua and Catherine are looking up at the top of a statue with angles of elevation of $29^\circ$29° and $43^\circ$43° respectively. If Catherine is standing $3$3 m closer to the statue than Joshua, what is the height $h$h of the statue? Using both triangles, we can find the distance from both Catherine and Joshua to the base of the statue in terms of $h$h.

$\text{Joshua to the statue}=\frac{h}{\tan29^\circ}$Joshua to the statue=htan29°

$\text{Catherine to the statue}=\frac{h}{\tan43^\circ}$Catherine to the statue=htan43°

Since Catherine is $3$3 m closer to the base of the statue, we know that:

$\frac{h}{\tan29^\circ}-\frac{h}{\tan43^\circ}=3$htan29°htan43°=3

Since this is now a simple equation involving only $h$h, we can solve for the height of the statue (rounding to two decimal places), which we find to be:

$h$h $=$= $\frac{3}{\frac{1}{\tan29^\circ}-\frac{1}{\tan43^\circ}}$31tan29°1tan43° $=$= $4.10$4.10

#### Practice questions

##### Question 1

Find the length of $AB$AB to two decimal places. ##### Question 2

$AB$AB is a tangent to a circle with centre $O$O.

$OB$OB is $20$20 cm long and cuts the circle at $C$C.

Find the length of $BC$BC to two decimal places. ##### Question 3

Find the size of angle $z$z.

Give your answer in degrees, minutes and seconds, rounding to the nearest second. ### Outcomes

#### MA5.2-13MG

applies trigonometry to solve problems, including problems involving bearings