4.5 Applications of trigonometric functions

4.5 Applications of trigonometric functions (EMCGK)

Area, sine and cosine rule

Area rule	Sine rule	Cosine rule
area $\triangle ABC = \frac{1}{2} bc \sin \hat{A}$	$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin \hat{C}}{c}$	$a^2 = b^2 + c^2 - 2 bc \cos A$
area $\triangle ABC = \frac{1}{2} ac \sin \hat{B}$	$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin \hat{C}}$	$b^2 = a^2 + c^2 - 2 ac \cos B$
area $\triangle ABC = \frac{1}{2} ab \sin \hat{C}$		$c^2 = a^2 + b^2 - 2 ab \cos \hat{C}$

How to determine which rule to use:

Area rule:
- no perpendicular height is given
Sine rule:
- no right angle is given
- two sides and an angle are given (not the included angle)
- two angles and a side are given
Cosine rule:
- no right angle is given
- two sides and the included angle are given
- three sides are given

Problems in two dimensions (EMCGM)

Worked example 13: Area, sine and cosine rule

Given $\triangle ABC$ with $a = \text{14}\text{ cm}, c = \text{17}\text{ cm}$ and $\hat{B} = \text{19} °$.

Calculate the following:

$b$
$\hat{C}$
Area $\triangle ABC$

Use the cosine rule to determine the length of $b$

\begin{align*} b^{2} &= a^{2} + c^{2} - 2ac \cos \hat{B} \\ &= (14)^{2} + (17)^{2} - 2(14)(17) \cos \text{19} ° \\ &= \text{34,93} \ldots \\ \therefore b &= \text{5,9}\text{ cm} \end{align*}

Use the sine rule to determine $\hat{C}$

\begin{align*} \frac{\sin \hat{C}}{c} &= \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{C}}{17} &= \frac{\sin \text{19} °}{\text{5,9}} \\ \sin \hat{C} &= \frac{17 \times \sin \text{19} °}{\text{5,9}} \\ \therefore \sin \hat{C} &= \text{0,938} \ldots \\ \text{First quadrant:} \enspace \hat{C} &= \text{69,7} ° \\ \text{Second quadrant:} \enspace \hat{C} &= \text{180} ° - \text{69,7} ° \\ &= \text{110,3} ° \end{align*}

From the diagram, we see that $\hat{C} > \text{90} °$, therefore $\hat{C} = \text{110,3} °$.

Calculate the area $\triangle ABC$

\begin{align*} \text{Area } \triangle ABC &= \frac{1}{2}ac \sin \hat{B} \\ &= \frac{1}{2} (14)(17) \sin \text{19} ° \\ &= \text{38,7}\text{ cm$^{2}$} \end{align*}

temp text

Worked example 14: Problem in two dimensions

In the figure below, $CD=BD=x$ and $B\hat{A}D = \theta$.

Show that $B{C}^{2}=2{x}^{2}\left(1+\sin\theta \right)$.

Consider the given information

Use the given information to determine as many of the unknown angles as possible.

\[\begin{array}{rll} CD &= BD = x & (\text{given} ) \\ B\hat{A}D &= \theta & (\text{given} ) \\ & \\ D\hat{B}A &= \text{90} ° & (\angle \text{ in semi circle}) \\ & \\ B\hat{D}A &= \text{180} ° - \text{90} ° - \theta & (\angle \text{s sum of } \triangle ABD) \\ &= \text{90} ° - \theta & \\ & \\ B\hat{D}C &= \text{90} ° + \theta & (\angle \text{s on a str. line} ) \end{array}\]

Determine the expression for $BC$

To derive the required expression, we need to write $BC$ in terms of $x$ and $\theta$.

In $\triangle CDB$, we can use the cosine rule to determine $BC$:

\begin{align*} B{C}^{2}& = C{D}^{2}+B{D}^{2}-2 \cdot CD \cdot BD \cdot \cos\left(B\hat{D}C\right) \\ & = {x}^{2}+{x}^{2}-2{x}^{2}\cos\left( \text{90} °+\theta \right) \\ & = 2{x}^{2} - 2{x}^{2}\left( - \sin\theta \right) \\ & = 2{x}^{2} + 2{x}^{2} \sin\theta \\ & = 2{x}^{2}\left(1 + \sin\theta \right) \end{align*}

Textbook Exercise 4.5

Find $A\hat{O}C$ in terms of $\theta$.

$BE$ is a diameter of semi-circle $BAE$. $O$ is the centre and therefore bisects $BE$. \[\begin{array}{rll} OA &= OE = OB &\text{ (equal radii)}\\ \therefore O \hat{A} E &= \theta &\text{ (opp. } \angle \text{'s of equal sides)}\\ \therefore A \hat{O} C &= \theta + \theta &\text{ (ext. } \angle \text{ of } \triangle OAE \text{ = sum int. opp.)}\\ &=2\theta & \end{array}\]

\[\begin{array}{rll} A \hat{O} C &= 2\theta & (\angle \text{ at centre } = 2 \angle \text{ at circum. }) \end{array}\]

In $\triangle ABE$, determine an expression for $\cos \theta$.

\begin{align*} B\hat{A}E &= \text{90} ° \quad (\angle \text{ in semi circle}) \\ \therefore \cos \theta &= \frac{AE}{BE} \end{align*}

In $\triangle ACE$, determine an expression for $\sin \theta$.

\begin{align*} A\hat{C}E &= \text{90} ° \quad (\text{given}) \\ \therefore \sin \theta &= \frac{CA}{AE} \end{align*}

In $\triangle ACO$, determine an expression for $\sin 2 \theta$.

\begin{align*} A\hat{C}O &= \text{90} ° \quad (\text{given}) \\ A\hat{O}C &= 2 \theta \quad (\text{proven}) \\ \therefore \sin 2 \theta &= \frac{CA}{AO} \end{align*}

Use the results from the previous questions to show that $\sin2\theta =2\sin\theta\cos \theta$.

\[\begin{array}{rll} \cos \theta &= \frac{AE}{BE} & \\ \therefore AE &= BE \cos \theta & \\ \sin \theta &= \frac{CA}{AE} & \\ \therefore CA &= AE \sin \theta & \\ &= (BE \cos \theta) \sin \theta & \\ BE &= 2(OE) & \text{ (radii)}\\ \therefore CA &= 2(OE) \sin \theta \cos \theta & \\ \therefore \frac{CA}{OE} &= 2 \sin \theta \cos \theta & \\ \text{but }\frac{CA}{OE} &= \frac{CA}{OA} = \sin 2\theta & \text{ (in }\triangle ACO \text{)}\\ \therefore \sin 2\theta &= 2 \sin \theta \cos \theta & \end{array}\]

$DC$ is a diameter of the circle with centre $O$ and radius $r$. $CA=r$, $AE=2DE$ and $D\hat{O}E=\theta$.

Show that $\cos\theta =\frac{1}{4}$.

\begin{align*} \text{In }\triangle DOE \text{, let }DE &= k \\ k^{2} &= r^{2} + r^{2} - 2r^{2} \cos \theta \\ &= 2r^{2} - 2r^{2} \cos \theta \ldots \ldots (1) \\ \text{In }\triangle AOE \text{, }AE&= 2k\\ (2k)^{2} &= (2r)^{2} + r^{2} - 2(2r)(r) \cos ( \text{180} ° - \theta) \\ \therefore 4k^{2} &= 4r^{2} + r^{2} + 4r^{2} \cos \theta\\ \therefore 4k^{2} &= 5r^{2} + 4r^{2} \cos \theta \ldots \ldots (2) \\ (1) \times 4 \text{: }4k^{2} &= 8r^{2} - 8r^{2} \cos \theta \ldots \ldots (3) \\ (3) - (2): \quad 0 &= 3r^{2} - 12r^{2} \cos \theta\\ \therefore 12r^{2}\cos\theta&= 3r^{2}\\ \cos \theta &= \frac{3r^{2}}{12r^{2}} \\ \therefore \cos \theta&= \frac{1}{4} \end{align*}

Show that the area of the cyclic quadrilateral is $DC \cdot DA \cdot \sin\hat{D}$.

\[\begin{array}{rll} \text{Connect } CA \text{ to construct }&\triangle ADC \text{ and }\triangle ABC &\\ \frac{BC}{CD} &=\frac{AD}{AB} \qquad \text{(given)} \\ \therefore BC \cdot AB &= AD \cdot CD \\ \text{Area }\triangle ADC &= \frac{1}{2} DC \cdot DA \sin \hat{D} & \\ \text{Area }\triangle ABC &= \frac{1}{2} AB \cdot BC \sin \hat{B} & \\ \hat{B} + \hat{D} &= \text{180} ° \qquad \text{ (opp. } \angle \text{s cyclic quad. supp.)}\\ \therefore \hat{B} &= \text{180} °- \hat{D} & \\ \therefore \sin \hat{B} &= \sin ( \text{180} ° - \hat{D}) = \sin \hat{D} & \\ \text{Area } ABCD &= \text{ area }\triangle ADC + \text{ area } \triangle ABC & \\ &= \frac{1}{2} DC \cdot DA \sin \hat{D} + \frac{1}{2} AB \cdot BC \sin \hat{B} & \\ &= \frac{1}{2} DC \cdot DA \sin \hat{D} + \frac{1}{2} DC \cdot DA \sin \hat{D} & \\ &= DC \cdot DA \sin \hat{D} & \end{array}\]

Write down two expressions for $CA^{2}$: one in terms of $\cos\hat{D}$ and one in terms of $\cos\hat{B}$.

\begin{align*} CA^{2} &= DC^{2} + DA^{2} - 2(DC)(DA) \cos \hat{D}\\ CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \cos \hat{B} \end{align*}

Show that $2C{A}^{2}=C{D}^{2}+D{A}^{2}+A{B}^{2}+B{C}^{2}$.

\begin{align*} CA^{2} &= DC^{2} + DA^{2} - 2(DC)(DA) \cos \hat{D}\\ CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \cos \hat{B}\\ &= AB^{2} + BC^{2} - 2(DC)(DA) \cos \hat{B} \quad (DC \cdot DA = AB \cdot BC) \\ 2CA^{2}& = CD^{2} + DA^{2} + AB^{2} + BC^{2} - 2(DC)(DA) \cos \hat{D} \\ & - 2(DC)(DA) \cos ( \text{180} ° -\hat{D})\\ &= CD^{2} + DA^{2} + AB^{2} + BC^{2} - 2(DC)(DA)\cos \hat{D} \\ & + 2(DC)(DA) \cos \hat{D}\\ &= CD^{2} + DA^{2} + AB^{2} + BC^{2} \end{align*}

Suppose that $BC=10$ units, $CD=15$ units, $AD=4$ units and $AB=6$ units. Calculate $C{A}^{2}$ (correct to one decimal place).

\begin{align*} 2CA^{2} &= (15)^{2} + (4)^{2} + (6)^{2} + (10)^{2} \\ &= 377\\ \therefore CA^{2} &= \text{188,5}\text{ units$^{2}$} \end{align*}

Find the angle $\hat{B}$. Hence, calculate the area of $ABCD$ (correct to one decimal place).

\begin{align*} CA^{2} &= AB^{2} + BC^{2} - 2(AB)(BC) \cos \hat{B} \\ \text{188,5}&= 6^{2} + 10^{2} - 2(6)(10) \cos \hat{B} \\ \text{188,5}&= 136 - 120 \cos \hat{B} \\ \cos \hat{B} &= - \text{0,4375} \\ \text{ref } \angle &= \text{64,05} ° \\ \therefore \hat{B} &= \text{180} ° - \text{64,05} ° \\ &= \text{115,94} ° \\ & \\ \text{Area } ABCD &= DC \times DA \sin ( \text{115,94} °)\\ &= 15 \times 4 \sin ( \text{115,94} °)\\ &= \sin ( \text{115,94} °)\\ & = \text{54,0}\text{ units$^{2}$} \end{align*}

Determine the minimum length of cable needed to connect $A$ and $C$ (to the nearest metre).

\begin{align*} \hat{B} &= \hat{D} = \text{90} ° \quad (\text{vertical towers}) \\ & \\ \text{In } \triangle ABP: \quad \frac{30}{AP} &= \sin \text{50} ° \\ AP &= \frac{30}{\sin \text{50} °} \\ &= \text{39,16}\text{ m} \\ & \\ \text{In } \triangle CDP: \quad \frac{27}{CP} &= \sin \text{35} ° \\ CP &= \frac{27}{\sin \text{35} °} \\ &= \text{47,07}\text{ m} \\ & \\ \text{In } \triangle APC: A\hat{P}C &= \text{180} ° - ( \text{50} ° + \text{35} ° ) \\ &= \text{95} ° \\ AC^{2} &= AP^{2} + PC^{2} - 2(AP)(PC) \cos A\hat{P}C \\ &= (\text{39,16})^{2} + (\text{47,07})^{2} - 2(\text{39,16})(\text{47,07}) \cos \text{95} ° \\ &= \text{4 070,39} \\ \therefore AC &= \text{63,80}\text{ m} \\ &\approx \text{64}\text{ m} \end{align*}

How far apart are the bases of the two towers (to the nearest metre)?

\begin{align*} \text{In } \triangle ABP: \quad \frac{BP}{\text{39,16}} &= \cos \text{50} ° \\ \therefore BP &= \text{39,16} \cos \text{50} ° \\ &= \text{25,17}\text{ m} \\ & \\ \text{In } \triangle CDP: \quad \frac{PD}{\text{47,07}} &= \cos \text{35} ° \\ \therefore PD &= \text{47,07} \cos \text{35} ° \\ &= \text{38,56}\text{ m} \\ & \\ BD &= BP + PD \\\ &= \text{25,17}\text{ m} + \text{38,56}\text{ m} \\ &= \text{63,7}\text{ m} \\ &\approx \text{64}\text{ m} \end{align*}

Problems in three dimensions (EMCGN)

Trigonometric formulae are useful for solving problems in two dimensions. However, in the real world all objects are three dimensional, so it is important that we extend the application of the area, sine and cosine formulae to three dimensional situations.

Drawing a three dimensional diagram is a crucial step in finding the solution to a problem. Interpreting the given information and sketching a three dimensional diagram are skills that need to be practised.

Worked example 15: Problems in three dimensions - height of a pole

$T$ is the top of a pole and its base, $F$, is in the same horizontal plane as the points $A$ and $B$. The angle of elevation measured from $B$ to $T$ is $\text{25} °$. $AB = \text{120}\text{ m}$, $F\hat{A}B = \text{40} ° \text{ and } F\hat{B}A = \text{30} °$.

Use the given information to calculate $h$, the height of the pole, to the nearest metre.

Draw a sketch

It is a challenge to analyze this situation when only a description is given. A diagram is very useful for representing three dimensional problems.

Draw a sketch using the given information. Indicate all right angles on the sketch, even those that do not look as though they are a $\text{90}$ ° angle (for example, $T\hat{F}B = \text{90} °$). It is also helpful to shade the horizontal plane ($\triangle FAB$), as it adds depth to the diagram and gives a better visualization of the situation.

Consider the given information

There are two triangles to consider: $\triangle FAB$ and $\triangle TFB$. We are required to find the length of $FT$, but the only information we have for $\triangle TFB$ is $F\hat{B}T = \text{25} °$.

Important: notice that $FB$ is a side of $\triangle TFB$ and it is also a side of $\triangle FAB$ - it is a link or bridge between the two triangles.

Therefore, to determine the height of the pole:

Write down all the given information on the sketch.
Use the information for $\triangle FAB$ to calculate $FB$.
Use $FB$ and the information for $\triangle TFB$ to calculate $FT$.

Determine the length of $FB$

We notice that $\triangle FAB$ does not have a right-angle, so we use the sine rule to determine $FB$.

\[\begin{array}{rll} \text{In } \triangle FAB: \quad \hat{F} &= \text{180} ° - \text{40} ° - \text{30} ° & (\angle \text{s sum of } \triangle FAB) \\ &= \text{110} ° & \\ & \\ \dfrac{FB}{ \sin \hat{A} } &= \dfrac{AB}{ \sin \hat{F} } & \\ & \\ \dfrac{FB}{ \sin \text{40} ° }&= \dfrac{120}{ \sin \text{110} ° } & \\ & \\ FB &= \dfrac{120 \times \sin \text{40} ° }{\sin \text{110} ° } & \\ & \\ \therefore FB &= \text{82,084} \ldots & \end{array}\]

Determine the length of $FT$

In $\triangle TFB$:

\[\begin{array}{rll} \hat{F} &= \text{90} ° & (\text{vertical pole}) \\ \hat{B} &= \text{25} ° & (\text{given}) \\ \tan \hat{B} &= \frac{FT}{FB} & \\ \tan \hat{B} \times FB &= FT & \\ \tan \text{25} ° \times \text{82,084} \ldots &= FT & \\ \therefore FT &= \text{38,246} \ldots & \\ &\approx \text{38}\text{ m} & (\text{to nearest metre}) \end{array}\]

Note: do not round off answers in the intermediate steps as this will affect the accuracy of the final answer. Always try to do the calculation in one complete step and only round off the final answer.

Write the final answer

The height of the pole, $h$, is $\text{38}\text{ m}$.

The calculation in the above worked example is only applicable for the specific numbers given. However, if we derived a general formula for this contextual situation, then it could be applied to any suitable set of numerical values.

Worked example 16: Problems in three dimensions - height of a building

$D$ is the top of a building of height $h$. The base of the building is at $A$ and $\triangle ABC$ lies on the ground (a horizontal plane). $BC=b$, $D\hat{B}A=\alpha$, $D\hat{B}C=\beta$ and $D\hat{C}B=\theta$.

Show that $h=\frac{b\sin\alpha\sin \theta }{\sin\left(\beta +\theta \right)}$.

Consider the given information

We know that $\triangle ABD$ is right-angled and we are required to find a formula to calculate the length of $AD$. In $\triangle BCD$, we are given two angles and a length of $BC$. We identify the side $BD$ as the link between $\triangle ABD$ and $\triangle BCD$.

Determine an expression for $BD$

$\triangle BCD$ does not have a right-angle but two angles and a side are given, so we use the sine rule to determine $BD$.

\[\begin{array}{rll} \text{In } \triangle BCD: \quad B\hat{D}C &= \text{180} ° - \left( \beta + \theta \right) & (\angle \text{s sum of } \triangle BCD) \\ \therefore \sin \left( \text{180} ° - \left( \beta + \theta \right) \right) &= \sin \left( \beta + \theta \right) & \\ \frac{BD}{\sin \theta } & = \frac{b}{\sin\left(B\hat{D}C\right)} \\ BD & = \frac{b\sin\theta }{\sin\left(B\hat{D}C\right)} \\ & = \frac{b\sin\theta }{\sin \left( \beta + \theta \right) } \end{array}\]

Determine an expression for $AD$

In $\triangle ABD$:

\[\begin{array}{rll} B\hat{A}D &= \text{90} ° & (\text{building vertical}) \\ D\hat{B}A &= \alpha & (\text{given}) \\ \frac{h}{BD}& = \sin \alpha \\ h & = BD \sin \alpha \\ \therefore h & = \frac{b \sin \alpha \sin\theta }{\sin \left( \beta + \theta \right) } \end{array}\]

Textbook Exercise 4.6

Find the height of the tower $BC$ in terms of $x$, $\tan\theta$ and $\cos\alpha$.

\begin{align*} \text{In } \triangle BCD, \quad \tan \theta &= \frac{BC}{BD} \\ \therefore BC &= BD \tan \theta\\ \text{In } \triangle ABD, \quad \frac{\sin \hat{A}}{BD} &= \frac{\sin \alpha}{x} \\ \therefore BD &= \frac{x \sin \hat{A}}{\sin \alpha}\\ \therefore BC &= \frac{x \sin \hat{A} \tan \theta}{\sin \alpha}\\ \hat{A} + \alpha + \alpha &= \text{180} ° \qquad \text{(}\angle \text{'s in }\triangle ABD\text{)}\\ \therefore \hat{A} &= \text{180} ° - 2\alpha\\ \therefore \sin \hat{A} & = \sin ( \text{180} ° - 2 \alpha) = \sin 2 \alpha\\ \therefore BC &= \frac{x \sin 2 \alpha \tan \theta}{\sin \alpha} \\ \text{ and } \sin 2 \alpha &= 2 \sin \alpha \cos \alpha \\ \therefore \frac{\sin 2 \alpha}{\sin \alpha} &= 2\cos \alpha\\ \therefore BC &= 2x \cos \alpha \tan \theta \end{align*}

Find $BC$ if we are given that $x= \text{140}\text{ m}$, $\alpha = \text{21} °$ and $\theta = \text{9} °$.

\begin{align*} BC &= 2(140)\cos \text{21} °\tan \text{9} ° = \text{41,40}\text{ m} \end{align*}

Use the given information to derive a general formula for $h$, the height of the mast.

\[\begin{array}{rll} \text{In } \triangle QAB: \quad \hat{Q} &= \text{180} ° - (x + y) & (\angle \text{s sum of } \triangle QAB) \\ \frac{QB}{ \sin \hat{A} } &= \frac{AB}{ \sin \hat{Q} } & \\ \frac{QB}{ \sin x }&= \frac{d}{ \sin \left( \text{180} ° - (x + y) \right) } & \\ \frac{QB}{ \sin x }&= \frac{d}{ \sin \left( x + y \right) } & \\ QB &= \frac{d \sin x}{\sin \left( x + y \right) } & \end{array}\]

In $\triangle PQB$:

\[\begin{array}{rll} \hat{Q} &= \text{90} ° & (\text{vertical mast}) \\ \hat{B} &= z & (\text{given}) \\ \tan \hat{B} &= \frac{QP}{QB} & \\ \tan \hat{B} \times QB &= QP & \\ \therefore QP & = \tan z \times QB & \\ \therefore h &= \tan z \times \frac{d \sin x}{\sin \left( x + y \right) } & \\ &= \frac{d \sin x \tan z}{\sin \left( x + y \right) } & \end{array}\]

If $d = \text{50}\text{ m}$, $x = \text{46} °$, $y = \text{15} °$ and $z = \text{20} °$, calculate $h$ (to the nearest metre).

\begin{align*} h &= \frac{d \sin x \tan z}{\sin \left( x + y \right) } \\ &=\frac{50 \sin \text{46} ° \tan \text{20} °}{\sin \left( \text{46} ° + \text{15} ° \right) } \\ &= \text{14,967} \ldots \\ &\approx \text{15}\text{ m} \qquad (\text{to nearest metre}) \end{align*}

Show that the height of the block of flats, $PR$, can be expressed as:

\[PR = q \tan \theta \left( \cos \alpha - \frac{\sqrt{3} \sin \alpha}{3} \right)\]

$QR$ is the link between $\triangle PQR$ and $\triangle SQR$.

\[\begin{array}{rll} \text{In } \triangle SQR: \quad \hat{S} &= \text{180} ° - ( \text{120} ° + \alpha) & (\angle \text{s sum of } \triangle SQR) \\ \frac{QR}{ \sin \hat{S} } &= \frac{q}{ \sin \hat{Q} } & \\ & \\ QR &= \dfrac{q \sin \left( \text{180} ° - ( \text{120} ° + \alpha) \right)}{ \sin \left( \text{120} ° \right) } & \\ & \\ &= \dfrac{q \sin \left( \text{60} ° - \alpha \right)}{ \sin \left( \text{180} ° - \text{60} ° \right)} & \\ & \\ &= \dfrac{q \left( \sin \text{60} ° \cos \alpha - \cos \text{60} ° \sin \alpha \right)}{ \sin \text{60} ° } & \end{array}\] \[\begin{array}{rll} &= \dfrac{q \left( \frac{\sqrt{3}}{2} \cos \alpha - \frac{1}{2} \sin \alpha \right)}{ \frac{\sqrt{3}}{2} } & \\ & \\ &= q \left( \dfrac{\sqrt{3} \cos \alpha -\sin \alpha }{2} \right) \times \dfrac{2}{\sqrt{3}} & \\ & \\ &= q \left( \cos \alpha - \dfrac{\sin \alpha}{\sqrt{3}} \right) & \\ & \\ &= q \left( \cos \alpha - \left(\dfrac{\sin \alpha}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \right) \right) & \\ & \\ &= q \left( \cos \alpha - \dfrac{\sqrt{3} \sin \alpha}{3} \right) & \\ \end{array}\] \[\begin{array}{rll} \text{In } \triangle PQR: \quad \hat{R} &= \text{90} ° & \\ \frac{PR}{QR} &= \tan \theta & \\ \therefore PR &= \tan \theta \times QR & \\ &= \tan \theta \times q \left( \cos \alpha - \frac{\sqrt{3} \sin \alpha}{3} \right) & \\ &= q \tan \theta \left( \cos \alpha - \frac{\sqrt{3} \sin \alpha}{3} \right) & \end{array}\]

If $SR = \text{35}\text{ m}, S\hat{R}Q = \text{16} °$ and $R\hat{Q}P = \text{30} °$, calculate $PR$ (correct to one decimal place).

\begin{align*} PR &= q \tan \theta \left( \cos \alpha - \frac{\sqrt{3} \sin \alpha}{3} \right) \\ &= 35 \tan \text{30} ° \left( \cos \text{16} ° - \frac{\sqrt{3} \sin \text{16} °}{3}\right) \\ &= \text{16,2}\text{ m} \end{align*}

Assuming each level is $\text{2,5}$ $\text{m}$ high, estimate the number of levels in the block of flats.

\begin{align*} \text{Approx. no. of levels }&= \frac{16}{\text{2,5}} \\ &= \text{6,4} \end{align*}

Therefore, there are approximately $\text{6}$ levels.

Show that the distance between the two ships is given by $SB = 400 \cos \beta$.

\[\begin{array}{rll} \text{In } & \triangle HSB: & \\ \quad HS &= HB = \text{200}\text{ m} & \text{(given)} \\ SB^{2} &= HS^{2} + HB^{2} - 2(HS)(HB) \cos S\hat{H}B & (\text{cosine rule}) \\ SB^{2} &= (200)^{2} + (200)^{2} - 2(200)(200) \cos S\hat{H}B & \\ &= \text{40 000}+\text{40 000} - \text{80 000} \cos \left( \text{180} ° - 2 \beta \right) & \\ &= \text{80 000} + \text{80 000} \cos \left( 2 \beta \right) & \\ &= \text{80 000}\left( 1 + \cos 2 \beta \right) & \\ &= \text{80 000}\left( 1 + 2 \cos^{2} \beta - 1 \right) & \\ &= \text{160 000} \cos^{2} \beta & \\ \therefore SB &= \text{400} \cos \beta & (\text{distance positive}) \end{array}\]

Show that the area of the sea included in $\triangle LSB$ is given by area $\triangle LSB = \text{2 000}\cos^{2} \alpha \sin \theta$.

\[\begin{array}{rll} \text{In } \triangle HSL &\text{ and } \triangle HBL: & \\ HS &= HB = \text{200}\text{ m} & (\text{given}) \\ HL &= HL & (\text{common side})\\ H\hat{L}S &= H\hat{L}B = \text{90} ° & (\text{vertical lighthouse}) \\ \therefore \triangle HSL &||| \triangle HBL & (\text{RHS}) \\ \therefore H\hat{B}L &= H\hat{S}L = \alpha & (\triangle HSL ||| \triangle HBL) \\ \text{Area} \triangle LSB &= \frac{1}{2} (LS)(LB) \sin \theta \\ \text{In } \triangle HSL: \quad \frac{LS}{HS} &= \cos \alpha & \\ \therefore LS &= HS \cos \alpha & \\ &= 200 \cos \alpha & \\ \text{In } \triangle HBL: \quad \frac{LB}{HB} &= \cos \alpha & \\ \therefore LB &= 200 \cos \alpha & \\ & \\ \text{Area} LSB &= \frac{1}{2} (LS)(LB) \sin \theta \\ &= \frac{1}{2} (200 \cos \alpha )(200 \cos \alpha ) \sin \theta \\ &= \text{20 000} \cos^{2} \alpha \sin \theta \end{array}\]

Calculate the triangular area of the sea if the angle of inclination from the ship to the top of the lighthouse is $\text{10} °$ and the angle between the direct lines from the base of the lighthouse to each ship is $\text{85} °$.

$\theta = \text{85} °$ and $\alpha = \text{10} °$.

\begin{align*} \text{Area }\triangle LSB &= \text{2 000} \cos^{2} \alpha \sin \theta \\ &= \text{2 000}(\cos \text{10} °)^{2} \sin \text{85} ° \\ &= \text{1 932,3}\text{ m$^{2}$} \end{align*}

A triangular look-out platform ($\triangle ABC$) is attached to a bridge that extends over a deep gorge. The vertical depth of the gorge, the distance from the edge of the look-out $C$ to the bottom of the gorge $D$, is $\text{13}$ $\text{m}$. The angle of depression from $A$ to $D$ is $\text{34}$ ° and from $B$ to $D$ is $\text{28}$ °. The angle at the edge of the platform, $\hat{C}$ is $\text{76}$ °.

Calculate the area of the look-out platform (to the nearest $\text{m$^{2}$}$).

\begin{align*} \text{Area } \triangle ABC &= \frac{1}{2} AC \cdot BC \sin \hat{C} \\ \text{In } \triangle ACD: \quad \frac{13}{AC} &= \tan \text{34} ° \\ \therefore AC &= \frac{13}{\tan \text{34} °} \\ &= \text{19,27} \ldots \\ \text{In } \triangle BCD: \quad \frac{13}{BC} &= \tan \text{28} ° \\ \therefore BC &= \frac{13}{\tan \text{28} °} \\ &= \text{24,44} \ldots \\ & \\ \text{Area } \triangle ABC &= \frac{1}{2} AC \cdot BC \sin \hat{C} \\ &= \frac{1}{2} (\text{19,27} \ldots)(\text{24,44} \ldots) \sin \text{76} ° \\ &= \text{228,57} \ldots \\ &\approx \text{229}\text{ m$^{2}$} \end{align*}

If the platform is constructed so that the two angles of depression, $C\hat{A}D$ and $C\hat{B}D$, are both equal to $\text{45}$ ° and the vertical depth of the gorge $CD = d$, $AB = x$ and $A\hat{C}B = \theta$, show that $\cos \theta = 1 - \frac{x^{2}}{2d^{2}}$.

\begin{align*} AB^{2} &= AC^{2} + BC^{2} - 2 \cdot AC \cdot BC \cos A\hat{C}B \\ \text{In } \triangle ACD: \quad \frac{d}{AC} &= \tan \text{45} ° \\ \therefore AC &= d \\ \text{And } BC &= d \\ \text{In } \triangle ABC: \quad x^{2} &= d^{2} + d^{2} - 2 \cdot d \cdot d \cos \theta \\ &= 2d^{2} - 2d^{2} \cos \theta \\ 2d^{2} \cos \theta &= 2d^{2} - x^{2} \\ \therefore \cos \theta &= \frac{2d^{2} - x^{2}} {2d^{2} } \\ &= 1 - \frac{x^{2}} {2d^{2} } \end{align*}

If $AB = \text{25}\text{ m}$ and $CD = \text{13}\text{ m}$, calculate $A\hat{C}B$ (to the nearest integer).

\begin{align*} \cos \theta &= 1 - \frac{x^{2}} {2d^{2} } \\ &= 1 - \frac{(25)^{2}} {2(13)^{2} } \\ \therefore \cos \theta&= - \text{0,849} \ldots \\ \text{ref } \angle &= \text{31,88} ° \\ \therefore \theta &= \text{180} ° - \text{31,88} ° \\ &= \text{148,12} ° \\ &\approx \text{148} ° \qquad (\text{nearest integer}) \end{align*}

4.4 Solving equations

Table of Contents

4.6 Summary

Area rule	Sine rule	Cosine rule
area \(\triangle ABC = \frac{1}{2} bc \sin \hat{A}\)	\(\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin \hat{C}}{c}\)	\(a^2 = b^2 + c^2 - 2 bc \cos A\)
area \(\triangle ABC = \frac{1}{2} ac \sin \hat{B}\)	\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin \hat{C}}\)	\(b^2 = a^2 + c^2 - 2 ac \cos B\)
area \(\triangle ABC = \frac{1}{2} ab \sin \hat{C}\)		\(c^2 = a^2 + b^2 - 2 ab \cos \hat{C}\)