topic badge
Middle Years

6.06 Problem solving

Lesson

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 $31$31 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 to two decimal places.

What is Mathspace

About Mathspace