 7.04 Further problem solving with trigonometry

Lesson

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

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.
When to use which?

Use the sine rule:

• When two angles and a corresponding side are known and you wish to find the other side
• When two sides and a corresponding angle are known and you wish to find the other angle

(Remember: while we require these in corresponding pairs we can always find the third angle if two are known)

Use the cosine rule:

• When two sides and the included angle are known and you wish to find the third side
• When all three sides are known and you wish to find an angle

Worked example

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

$\cos B=\frac{a^2+c^2-b^2}{2ac}$cosB=a2+c2b22ac

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 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°​ $d$d $=$= $116.36$116.36 m (To 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° $height$height $=$= $94.2265$94.2265 ... m

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

$\triangle ABC$ABC consists of angles $A$A, $B$B and $C$C which appear opposite sides $a$a, $b$b and $c$c respectively. Consider the case where the measures of $a$a, $c$c and $A$A are given.

1. Which of the following is given?

$SSA$SSA: Two sides and an angle

A

$SAS$SAS: Two sides and the included angle

B

$SAA$SAA: two angles and a side

C

$ASA$ASA: two angles and the side between them

D

$SSS$SSS: Three sides

E

$SSA$SSA: Two sides and an angle

A

$SAS$SAS: Two sides and the included angle

B

$SAA$SAA: two angles and a side

C

$ASA$ASA: two angles and the side between them

D

$SSS$SSS: Three sides

E
2. Which law should be used to start solving the triangle?

the law of sines

A

the law of cosines

B

the law of sines

A

the law of cosines

B

Question 2

Mae observes a tower at an angle of elevation of $12^\circ$12°. The tower is perpendicular to the ground.

Walking $67$67 m towards the tower, she finds that the angle of elevation increases to $35^\circ$35°. 1. Calculate the angle $\angle ADB$ADB.

2. Find the length of the side $a$a.

3. Using the rounded value of the previous part, evaluate the height $h$h, of the tower.

Question 3

Dave leaves town along a road on a bearing of $169^\circ$169° and travels $26$26 km. Maria leaves the same town on another road with a bearing of $289^\circ$289° and travels $9$9 km.

Find the distance between them, $x$x.

Outcomes

4.2.1.3

establish and use the sine (ambiguous case is required) and cosine rules and the formula 𝐴rea=1/2 bc sin𝐴 for the area of a triangle

4.2.1.4

construct mathematical models using the sine and cosine rules in two- and three-dimensional contexts (including bearings in two-dimensional context) and use the model to solve problems; verify and evaluate the usefulness of the model using qualitative statements and quantitative analysis