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?

Goal: Find length $AB$AB.

In triangle $ABC$ABC 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, angle B is $180-69-47=64$180−69−47=64° (angles in a triangle sum to $180$180)

Using the sine rule:

$\frac{b}{\sin B}$`bsinB`	$=$=	$\frac{d}{\sin D}$`dsinD`	$d$`d` is length $BC$`BC`, named d because it is opposite angle $D$`D`
$\frac{143}{\sin64^\circ}$143`sin`64°	$=$=	$\frac{d}{\sin47^\circ}$`dsin`47°	then we can rearrange to find length $d$`d`
$\frac{143\sin47^\circ}{\sin64^\circ}$143`sin`47°`sin`64°	$=$=	$d$`d`
$d$`d`	$=$=	$116.36$116.36 m

Now we are only halfway through. We still need to find the height of the cliff.

Using tan we can complete the question

$\tan39^\circ$`tan`39°	$=$=	$\frac{height}{116.36}$`height`116.36
$height$`height`	$=$=	$116.36\times\tan39^\circ$116.36×`tan`39°
$height$`height`	$=$=	$94.2265$94.2265 ...

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

Using the sine rule

The sine rule is useful when you want to find:

a side length if you know 2 angles and another side
an angle size if you know 1 angle and 2 corresponding side lengths

The best way to start problems involving applications of the sine rule is to label the angles of the triangle, label the corresponding sides and then use the most appropriate version of the sine rule.

The sine rule

$\frac{a}{\sin A}=\frac{b}{\sin B}$asinA=bsinB =$\frac{c}{\sin C}$csinC,

we can also use the alternative form

$\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa=sinBb = $\frac{\sin C}{c}$sinCc

depending on what we are trying to use the law for.

The sine rule shows: that the lengths of the sides in a triangle are proportional to the

sines of the measures of the angles opposite them

Here are some worked examples.

Question 1

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°.

A tower is depicted as a vertical line segment $CD$`CD` with point $C$ as the base and point $D$`D` as the top of the tower. Point $C$`C` is connected with a horizontal line segment to points $A$`A`, and $B$`B` on its left. $B$`B` lies on $AC$`AC`. Two line segments connect $A$`A` and $B$`B` to $D$`D`. Triangles $\triangle ACD$△`ACD`, $\triangle BCD$△`BCD`, and $\triangle ABD$△`ABD` are formed by the line segments and the tower. $CD$`CD` is labeled as $h$`h` , indicating its height. $BD$`BD` is labeled as $a$`a`, indicating its length. $AB$`AB` measures $67$ m as labeled. The angle formed between $AD$`AD` and $AC$`AC` is marked by an arc marking and measures $12^\circ$12°. The angle formed between $BD$`BD` and $BC$`BC` is also marked by an arc marking and measures $35^\circ$35°.

Calculate the angle $\angle ADB$∠ADB.
Find the length of the side $a$a.

Round your answer to two decimal places.
Using the rounded value of the previous part, evaluate the height $h$h, of the tower.

Round your answer to one decimal place.

Question 3

$Lucy$ travelled on a bearing of $39^\circ$39° from point A to point B. $She$ then travelled on a bearing of $159^\circ$159° for $17$17 km to point C which is east of point A.

Find the size of the angle $\angle CAB$∠CAB.
Find the size of the angle $\angle ABC$∠ABC
Solve for $x$x, the distance $Lucy$ would have to travel due east to return to point A.

Give your answer to two decimal places.

Outcomes

11U.D.1.6

Pose problems involving right triangles and oblique triangles in two dimensional settings, and solve these and other such problems using the primary trigonometric ratios, the cosine law, and the sine law (including the ambiguous case)

Applications of the sine law

Example