topic badge

8.06 Mixed applications of 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. 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+b22abcosC

 

Alternatively, for finding an unknown angle:

 

$\cos C=\frac{a^2+b^2-c^2}{2ab}$cosC=a2+b2c22ab.

Area of a triangle

For a triangle with side lengths $a$a and $b$b and an included angle $C$C, then the area of the triangle is given by:

$Area=\frac{1}{2}ab\sin C$Area=12absinC

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 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+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+182230422×150×182
  $=$= $\frac{-36792}{54600}$3679254600
$B$B $=$= $\cos^{-1}\left(\frac{-36792}{54600}\right)$cos1(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

$\triangle ABC$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
  2. Which law should be used to start solving the triangle?

    the law of sines

    A

    the law of cosines

    B

Question 2

A helicopter is flying at an altitude of $136$136 metres, at an angle of depression of $35$35° to its landing pad. What is the distance $d$d between the helicopter and the landing pad?

Round your answer to the nearest whole number.

A right triangle illustrates the scenario of a helicopter (positioned in the upper left) in flight and its landing pad (positioned in the lower right). The hypotenuse, representing the direct slant distance from the helicopter to the landing pad, is labeled d. The angle at the helicopter’s location, denoted as 35 degrees is the angle of depression towards the landing pad. The side opposite the angle of 35 degrees, indicating the helicopter’s altitude, measures 136 meters.

question 3

A man stands at point $A$A looking at the top of two poles. Pole $1$1 has height $8$8 m and angle of elevation $34^\circ$34° and Pole $2$2 has height $25$25 m and angle of elevation $57^\circ$57°. The man wishes to find the distance between the two poles.

Two poles are shown and labeled as Pole 1 and Pole 2. A point on the ground is also labeled as A. Point A is on the left side, Pole 2 is on the right side and Pole 1 is in between. The angle of elevation between the point on the ground to the top of the two poles forms two triangles where the smaller triangle is created inside the bigger triangle.

For the bigger triangle, the line of elevation from point A on the ground to the top of Pole 2 acts as the hypotenuse of the triangle. The angle between the hypotenuse and the ground is the angle of elevation. The horizontal side adjacent to this angle, which is labeled as y, represents the distance from A to the base of Pole 2 labeled as C. Opposite to the angle is the height of Pole 2 which measures 25 m, which is also the vertical side of the triangle.

For the smaller triangle, the line of elevation from point A on the ground to the top of Pole 1 acts as the hypotenuse of the triangle. The angle between the hypotenuse and the ground is the angle of elevation. The horizontal side adjacent to this angle, which is labeled as $x$x, represents by the distance from A to the base of Pole 1 labeled as B. Opposite to the angle is the height of Pole 2 which measures 8 m, which is also the vertical side of the triangle.

  1. Find $x$x, the distance from $A$A to $B$B in metres.

    Round your answer to two decimal places.

  2. Now find $y$y, the distance from $A$A to $C$C in metres.

    Round your answer to two decimal places.

  3. Hence find $BC$BC, the distance between the two poles in metres.

    Round your answer to one decimal place.

QUESTION 4

Calculate the length of $y$y in metres.

Round your answer to one decimal place.

question 5

Consider the following diagram.

  1. Calculate the direct distance, $x$x km, from $P$P to $Q$Q.

    Round your answer to two decimal places.

  2. Calculate the area $A$A, in km2, enclosed by triangle $POQ$POQ.

    Round your answer to two decimal places.

  3. Calculate the bearing $\alpha$α of $Q$Q from $P$P.

    Round your answer to the nearest minute.

Question 6

Farmer Joe has a trapezoidal shaped paddock. He is trying to calculate the area and has some of the measurements of the paddock.

  1. What is the area of his paddock?

    Round your answer to two decimal places.

  2. For the neighbouring paddock, Farmer Joe’s son Jack decides he can determine the area with fewer measurements. He measures the following:

    What is the area of the neighbouring paddock?
    Round your answer to two decimal places.

Outcomes

ACMGM037

solve practical problems involving the trigonometry of right-angled and non-right-angled triangles, including problems involving angles of elevation and depression and the use of bearings in navigation

What is Mathspace

About Mathspace