Victorian Curriculum Year 10A - 2020 Edition
7.08 Applications
Lesson

Consider $A,B,C$A,B,C as the vertices of a triangle and $a,b,c$a,b,c as the side lengths opposite to each angle respectively. If this triangle is a non-right-angled triangle then we can use the following rules. If the triangle is right-angled then we can use Pythagoras' theorem and the trigonometric ratios.

### Applications

Both the sine rule and the cosine rule can be used to find missing angles and sides in a non-right-angled triangle. Consider

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

If three of the four values are known, then we can find the missing angle or side.The important point to note is that we need to be able to relate angles to their opposite sides.

For the cosine rule we have

$c^2=a^2+b^2-2ab\cos C$c2=a2+b22abcosC

For the cosine rule we need to one angle and two sides.

We are now going to use these rules to find these unknown quantities in real world contexts. A good way to begin any question involving a triangle is to label the angles and their corresponding sides using the letters from the formula above. After that is done correctly we can use the rule that is most convenient.

#### Exploration

There are three flight paths joining Adelaide, Sydney, and Brisbane to one another. The flight path joining Adelaide and Sydney is $1601$1601 km long, and the flight path joining Sydney and Brisbane is $732$732 km long. The angle between the two flight paths meeting at Adelaide is $20^\circ$20°.
An aircraft will travel from Sydney to Adelaide via Brisbane. Find $\theta$θ, the angle between the two flight paths meeting at Brisbane, to the nearest degree.

The line segments between each city forms a triangle. We know the lengths of two sides and one angle, so we can use the sine rule ($\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa=sinBb) to find the size of the angle $\theta$θ. Here is the triangle without the underlying map, with sides and angles labelled in a corresponding way:

 $\frac{\sin A}{a}$sinAa​ $=$= $\frac{\sin B}{b}$sinBb​ Write down the sine rule. $\frac{\sin\theta}{1160}$sinθ1160​ $=$= $\frac{\sin20^\circ}{732}$sin20°732​ Substitute in the known values. $\sin\theta$sinθ $=$= $\frac{1160\sin20^\circ}{732}$1160sin20°732​ Multiplying both sides by $1160$1160. $\theta$θ $=$= $\sin^{-1}\left(\frac{1160\sin20^\circ}{732}\right)$sin−1(1160sin20°732​) Take the inverse sine of both sides. $\theta$θ $=$= $33^\circ$33° Calculate the answer, rounded to nearest degree.

To find the distance between Brisbane and Adelaide we need the side length opposite the one angle we don't have.

However we can find the final angle using the angle sum of a triangle. $180^\circ-\left(20^\circ+33^\circ\right)=127^\circ$180°(20°+33°)=127°.

We can find the distance using the cosine rule or the sine rule. To find it using the cosine rule we do the following:

 $c^2$c2 $=$= $a^2+b^2-2ab\cos C$a2+b2−2abcosC $c$c $=$= $\sqrt{a^2+b^2-2ab\cos C}$√a2+b2−2abcosC Take square root of both sides. $=$= $\sqrt{732^2+1160^2-2\times732\times1160\cos127^\circ}$√7322+11602−2×732×1160cos127° Substitute side lengths and angle into equation. $=$= $1703.951$1703.951$\cdots$⋯ Evaluate using your calculator. $=$= $1704$1704 km Rounded to the nearest km.

So the distance between Adelaide and Brisbane is $1704$1704 km.

Remember!

We can use the following formulas for non-right-angled triangles:

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

$c^2=a^2+b^2-2ab\cos C$c2=a2+b22abcosC

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

For right-angled triangles we can use the trigonometric ratios and Pythagoras' theorem $a^2+b^2=c^2$a2+b2=c2.

#### Practice questions

##### question 1

Find the length of the unknown side, x, in the given trapezium.

Give your answer correct to $2$2 decimal places.

##### QUESTION 2

Find the value of angle $w$w in degrees.

##### question 3

$VUTR$VUTR is a rhombus with perimeter $76$76 cm. The length of diagonal $RU$RU is $34$34 cm.

1. First find the length of $VR$VR.

2. Then find the length of $RW$RW.

3. If the length of $VW$VW is $x$x cm, find $x$x correct to 2 decimal places.

4. Hence, what is the length of the other diagonal $VT$VT correct to 2 decimal places.

### Outcomes

#### VCMMG367 (10a)

Establish the sine, cosine and area rules for any triangle and solve related problems.