8.06 Mixed applications of trigonometry

We have looked at using both the sine and cosine rules for finding unknown sides and angles in problems involving non-right-angled triangles. We have also looked at finding the area of a non-right-angled triangle.

For a triangle with sides $a$a, $b$b, and $c$c, with corresponding angles $A$A, $B$B, and $C$C:

The sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$asinA​=bsinB​=csinC​   Alternatively, for finding an unknown angle:   $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$sinAa​=sinBb​=sinCc​	The cosine rule $c^2=a^2+b^2-2ab\cos C$c2=a2+b2−2abcosC   Alternatively, for finding an unknown angle:   $\cos C=\frac{a^2+b^2-c^2}{2ab}$cosC=a2+b2−c22ab​.

We will now look at application problems that will require us to choose whether to use the sine rule or the cosine rule, or indeed other aspects of trigonometry. 

 

Worked examples

Example 1

Scientists can use a set of footprints to calculate an animal's step angle, which is a measure of walking efficiency. The closer the step angle is to $180^\circ$180°, the more efficiently the animal walked.

The angle marked here highlights the step angle.

If a particular set of footprints has length $AC=304$AC=304 cm, length $BC=150$BC=150 cm and length $AB=182$AB=182 cm, find the step angle.

Think: We have 3 side lengths, and require a missing angle, so we will use the cosine rule.

If angle $B$B is the step angle we are trying to find, then:

$\cos B=\frac{a^2+c^2-b^2}{2ac}$cosB=a2+c2−b22ac​

Do: Then fill in all the values we know,

$\cos\left(B\right)$cos(B)	$=$=	$\frac{150^2+182^2-304^2}{2\times150\times182}$1502+1822−30422×150×182​
	$=$=	$\frac{-36792}{54600}$−3679254600​
$B$B	$=$=	$\cos^{-1}\left(\frac{-36792}{54600}\right)$cos−1(−3679254600​)
	$=$=	$132.36^\circ$132.36° (to $2$2 decimal places)

Reflect: Notice that the cosine ratio of the angle is negative. This indicates that the angle will be greater than $90^\circ$90°.

 

Example 2

At a picnic by the river two children wondered how tall the cliff opposite the river was. By taking some measurements, like the angles and distances shown, can you help them find the height of the cliff?

Think: The goal is to find length $AB$AB. What do we know:

• Triangle $ABC$ABC is right-angled but we only have angle measurements. Before we can use trigonometry to find length $AB$AB we will need a side length. Length $BC$BC can be found using triangle $BCD$BCD.
• In triangle $BCD$BCD, we have two angles and a side. We do not have the angle corresponding to the known side but we can find it: $B=180^\circ-69^\circ-47^\circ=64^\circ$B=180°−69°−47°=64°(angles in a triangle sum to $180^\circ$180°)
• Hence, for triangle $BCD$BCD we know two angles and a corresponding side, we can use the sine rule to find side $CB$CB.

Do:

Using sine rule:

$\frac{d}{\sin D}$dsinD​	$=$=	$\frac{b}{\sin B}$bsinB​	$d$d is length $BC$BC, named d because it is opposite angle $D$D.
$\frac{d}{\sin47^\circ}$dsin47°​	$=$=	$\frac{143}{\sin64^\circ}$143sin64°​	Then we can rearrange to find length $d$d.
$d$d	$=$=	$\frac{143\sin47^\circ}{\sin64^\circ}$143sin47°sin64°​
	$=$=	$116.36$116.36 m (to $2$2 decimal places)

Now we have enough information to find the height of the cliff.

Using right-angled trigonometry we can complete the question:

$\tan\theta$tanθ	$=$=	$\frac{O}{A}$OA​
$\tan39^\circ$tan39°	$=$=	$\frac{height}{116.36}$height116.36​
$height$height	$=$=	$116.36\times\tan39^\circ$116.36×tan39°
	$=$=	$94.23$94.23 m (to $2$2 decimal places)

So the cliff is approximately $94$94 m tall.

Reflect: For practical problems reflect on whether your answer seems realistic. For this case, this is quite a tall cliff but the answer is certainly possible, we did not get a negative height or unrealistically short or tall cliff.

 

Practice questions

Question 1

Question 2

question 3

QUESTION 4

question 5

Question 6