L.C. MATHS

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Past Papers

## Trigonometry

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Pythagoras' Theorem

Trigonometric Ratios

Area of a Triangle

Sine/Cosine Rules

## 2022 Paper 2 Question 5

(a) An equilateral triangle $$PQR$$ has sides of length $$8\mbox{ cm}$$.

(i) Write down the size of the angle $$\angle PQR$$.

(ii) Show that the area of the triangle $$PQR$$ is $$16\sqrt{3}\mbox{ cm}^2$$.

(iii) Hence, or otherwise, find the perpendicular height of the triangle $$PQR$$, taking $$PQ$$ as
the base. Give your answer in the form $$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

(i) $$60^{\circ}$$

(ii) The answer is already in the question!

(iii) $$4\sqrt{3}\mbox{ cm}$$

Solution

(i) $$60^{\circ}$$

(ii)

\begin{align}\mbox{Area}&=\frac{1}{2}ab\sin C\\&=\frac{1}{2}(8)(8)\sin60^{\circ}\\&=32\left(\frac{\sqrt{3}}{2}\right)\\&=16\sqrt{3}\mbox{ cm}^2\end{align}

(iii)

\begin{align}16\sqrt{3}=\frac{1}{2}bh\end{align}

\begin{align}\downarrow\end{align}

\begin{align}16\sqrt{3}=\frac{1}{2}(8)h\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\frac{2(16\sqrt{3})}{8}\\&=4\sqrt{3}\mbox{ cm}\end{align}

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(b) $$GHK$$ is a right-angled triangle.
$$|\angle GHK|=90^{\circ}$$
$$|GH|=12\mbox{ cm}$$
and $$|GK|=30\mbox{ cm}$$.

Using the theorem of Pythagoras, find the distance $$|HK|$$.
Give your answer correct to $$1$$ decimal place.

$$27.5\mbox{ cm}$$

Solution

\begin{align}|GK|^2=|GH|^2+|HK|^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|HK|&=\sqrt{|GK|^2-|GH|^2}\\&=\sqrt{30^2-12^2}\\&\approx27.5\mbox{ cm}\end{align}

Video Walkthrough

## 2022 Paper 2 Question 8(b) &(d)

The top of a particular lighthouse is in the shape of a hemisphere on top of a cylinder.
The hemisphere and the cylinder both have a radius of $$3\mbox{ m}$$.

(b) The diagram on the right below shows part of the base of the lighthouse (not to scale).
It is in the shape of a cone of radius $$7.5\mbox{ m}$$, from which the top part has been removed, leaving a horizontal circle of radius $$3\mbox{ m}$$.

(i) The height of the cone before the top part is removed is $$47\mbox{ m}$$.
Work out the size of the angle at the base of the cone, marked $$A$$ in the diagram above.
Give your answer correct to the nearest degree.

(ii) Find the distance marked $$k$$ on the diagram, the height after the top part is removed.

(i) $$81^{\circ}$$

(ii) $$28.2\mbox{ m}$$

Solution

(i)

\begin{align}A&=\tan^{-1}\left(\frac{47}{7.5}\right)\\&\approx81^{\circ}\end{align}

(ii)

\begin{align}\frac{x}{47}=\frac{3}{7.5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{(3)(47)}{7.5}\\&=18.8\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k&=47-x\\&=47-18.8\\&=28.2\mbox{ m}\end{align}

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(d) The top of the Fastnet lighthouse, $$F$$, is $$49\mbox{ m}$$ above sea level.

The angle of elevation of the top of the lighthouse from a ship $$S$$ is $$1.2^{\circ}$$, as shown in the
diagram below (not to scale).

Find the horizontal distance marked $$d$$ below, from the ship to the base of the lighthouse.
Give your answer in kilometres, correct to $$2$$ decimal places.

$$2.34\mbox{ km}$$

Solution

\begin{align}\tan1.2^{\circ}=\frac{49}{d}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}d&=\frac{49}{\tan1.2^{\circ}}\\&\approx2.34\mbox{ km}\end{align}

Video Walkthrough

## 2022 Paper 2 Question 9(c) & (e)

Seán has built a shed. The diagrams below show the dimensions of Seán’s shed.
The shed is in the shape of a prism. Its front face is in the shape of a triangle on top of a rectangle.
Its highest point is directly above the centre of its base.

(c) Use the theorem of Pythagoras to find the length of the distance marked $$d$$ in the diagram below, the slant length of the roof. Give your answer in metres, correct to $$1$$ decimal place.

$$6.2\mbox{ m}$$

Solution

\begin{align}d&=\sqrt{6^2+1.5^2}\\&\approx6.2\mbox{ m}\end{align}

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(e) The diagram below shows part of the roof of a smaller shed (not to scale).
Some measurements are marked on the diagram.

(i) Show that $$|BC|=4.65\mbox{ m}$$, correct to $$2$$ decimal places.

(ii) Find $$|\angle ACB|$$, the angle that the roof makes at the point $$C$$.
Give your answer correct to the nearest degree.
Remember that $$|BC|=4.65\mbox{ m}$$, correct to $$2$$ decimal places.

(i) The answer is already in the question!

(ii) $$19^{\circ}$$

Solution

(i)

\begin{align}|BC|&=\sqrt{|AB|^2+|AC|^2-2|AB||AC|\cos|\angle BAC|}\\&=\sqrt{3^2+7^2-2(3)(7)\cos(30^{\circ}}\\&\approx4.65\mbox{ m}\end{align}

as required.

(ii)

\begin{align}|\angle ACB|&=\sin^{-1}\left(\frac{3\sin30^{\circ}}{4.65}\right)\\&\approx19^{\circ}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 2(b)

(b) The figure $$ABCDE$$ shown in the diagram consists of a large square $$ACDE$$ standing on the diagonal $$[AC]$$ of a smaller square $$ABCF$$.
The smaller square has a side length of $$2\mbox{cm}$$.
Find the area and perimeter of the figure $$ABCDE$$.
Give your answer for the perimeter correct to two decimal places.

$$\mbox{Area}=10\mbox{ cm}^2$$ and $$\mbox{Perimeter}=12.49\mbox{ cm}$$

Solution

\begin{align}|AC|&=\sqrt{|AB|^2+|BC|^2}\\&=\sqrt{2^2+2^2}\\&\sqrt{8}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{Area}&=(\sqrt{8})^2+\frac{1}{2}(2\times2)\\&=10\mbox{ cm}^2\end{align}

and

\begin{align}\mbox{Perimeter}&=\sqrt{8}+\sqrt{8}+\sqrt{8}+2+2\\&\approx12.49\mbox{ cm}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 8(c)

(c) The buoy is situated at B, $$43a\mbox{ m}$$ from the pier $$P$$ and $$26\mbox{ m}$$ from the point $$Q$$, as shown in the
diagram above.
Find $$|\angle QBP|$$, the angle at the buoy shown on the diagram.
Give your answer correct to two decimal places.

$$118.74^{\circ}$$

Solution

\begin{align}60^2=43^2+26^2-2(43)(26)\cos|\angle QBP|\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|\angle QBP|&=\cos^{-1}\left(\frac{43^2+26^2-60^2}{2(43)(26)}\right)\\&\approx118.74^{\circ}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 9

The line segment $$[SE]$$, shown below, represents an airport runway.
The point $$S$$ and the point $$E$$ represent the start and end points of the runway respectively.
The diagram is drawn to a scale of $$1\mbox{ unit}=250\mbox{ metres}$$.

(a)

(i) Find the length of the runway. Give your answer in $$\mbox{km}$$.

(ii) An aircraft starts at the point S and travels $$1250\mbox{ m}$$ to a point $$L$$ where it lifts off.
From the point $$L$$, the aircraft makes an angle of $$14^{\circ}$$ with the ground, $$[LE]$$.
The aeroplane travels along this path until it is directly above $$E$$.
Plot and label the lift-off point $$L$$, and draw the aircraft’s flight path for this part of its journey, on the diagram above.

(iii) Find the total distance the aeroplane has travelled when it is directly above $$E$$.
Give your answer, in metres, correct to the nearest metre.

(i) $$2.5\mbox{ km}$$

(ii)

(iii) $$2{,}538\mbox{ m}$$

Solution

(i)

\begin{align}l&=250\times 10\\&=2{,}500\mbox{ m}\\&=2.5\mbox{ km}\end{align}

(ii)

(iii)

\begin{align}d&=1{,}250+\frac{1{,}250}{\cos 14^{\circ}}\\&\approx2{,}538\mbox{ m}\end{align}

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(b) The aircraft flies from airport $$A$$ to airport $$B$$, and then on to airport $$C$$, at the same altitude.
The pilot records the flight summary on the given diagram.

(i) Find the distance from airport $$B$$ to airport $$C$$. Give your answer correct to the nearest $$\mbox{km}$$.

(ii) When the plane was directly over airport $$C$$, the pilot was instructed to “circle” until a runway was available. She therefore flew $$10\mbox{ km}$$ away from $$C$$ before turning and flying along a circular arc of $$70^{\circ}$$ and then returning to the airport.
This path was all flown at the same altitude.
Find the total distance travelled.
Give your answer, in $$\mbox{km}$$, correct to two decimal places.

(i) $$324\mbox{ km}$$

(ii) $$32.22\mbox{ km}$$

Solution

(i)

\begin{align}\frac{|BC|}{\sin47^{\circ}}=\frac{260}{\sin36^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|BC|&=\frac{260(\sin47^{\circ})}{\sin36^{\circ}}\\&\approx324\mbox{ km}\end{align}

(ii)

\begin{align}d&=10+10+2\pi(10)\left(\frac{70^{\circ}}{360^{\circ}}\right)\\&\approx32.22\mbox{ km}\end{align}

Video Walkthrough

## 2020 Paper 2 Question 5

(a) Two swimmers $$A$$ and $$B$$ stand at the same point $$X$$, on one shore of a long, still rectangular shaped lake that is $$100\mbox{ m}$$ wide, as shown below. (Diagram not to scale.)
Both swim to the opposite side of the lake.

$$122\mbox{ m}$$

Solution

\begin{align}\sin55^{\circ}=\frac{100}{d}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}d&=\frac{100}{\sin55^{\circ}}\\&\approx122\mbox{ m}\end{align}

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(b) Swimmer $$B$$ swims to the right and travels a distance of $$200\mbox{ m}$$ to reach the other side, making an angle of $$\theta$$ degrees with the bank on the other side of the lake, as shown.
Find the value of $$\theta$$.

$$30^{\circ}$$

Solution

\begin{align}\sin \theta &=\frac{100}{200}\\&=\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta &=\sin^{-1}\left(\frac{1}{2}\right)\\&=30^{\circ}\end{align}

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(c) The next day the swimmers again swim to the opposite side of the lake but in slightly different directions.
Swimmer $$A$$ swims to the left, making an angle of $$45^{\circ}$$ with the side of the lake and travels $$141.4$$ metres as shown.
Swimmer $$B$$ swims to the right, making an angle of $$40^{\circ}$$ with the side of the lake and travels $$155.6$$ metres as shown.
Find $$d$$, the distance both swimmers are apart when they reach the opposite side of the lake.
Give you answer correct to the nearest metre.

$$219\mbox{ m}$$

Solution

\begin{align}d&=\sqrt{141.4^2+155.6^2-2(141.4)(155.6)\cos95^{\circ}}\\&\approx219\mbox{ m}\end{align}

Video Walkthrough

## 2020 Paper 2 Question 7(a) & (c)

A vertical mobile phone mast, $$[DC]$$, of height $$h\mbox{ m}$$, is secured with two cables: $$[AC]$$ of length $$x\mbox{ m}$$,
and $$[BC]$$ of length $$y\mbox{ m}$$, as shown in the diagram.
The angle of elevation to the top of the mast from $$A$$ is $$30^{\circ}$$ and from $$B$$ is $$45^{\circ}$$.

(a)

(i) Explain why $$|\angle BCA|$$ is $$105^{\circ}$$.

(ii) The horizontal distance from $$A$$ to $$B$$ is $$100\mbox{ m}$$.
Use the triangle $$ABC$$ to find the length of $$y$$.
Give your answer correct to one decimal place.

(iii) Using your answer to Part (a)(ii) or otherwise, find the value of $$h$$ and the value of $$x$$.
Give your answers correct to $$1$$ decimal place.

(i) The answer is already in the question!

(ii) $$y=51.8\mbox{ m}$$

(iii) $$h=36.6\mbox{ m}$$ and $$x=73.2\mbox{ m}$$

Solution

(i)

\begin{align}|\angle BCA|&=|\angle ACD|+|\angle DCB|\\&=60^{\circ}+45^{\circ}\\&=10^{\circ}\end{align}

as required.

(ii)

\begin{align}\frac{y}{\sin30^{\circ}}=\frac{100}{\sin105^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{100\sin30^{\circ}}{\sin105^{\circ}}\\&\approx51.8\mbox{ m}\end{align}

(iii)

\begin{align}\sin45^{\circ}&=\frac{h}{51.8}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=51.8\sin45^{\circ}\\&\approx36.6\mbox{ m}\end{align}

and

\begin{align}\sin30^{\circ}=\frac{36.6}{x}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{36.6}{\sin30^{\circ}}\\&\approx73.2\mbox{ m}\end{align}

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(c)

(i) The mast can provide a strong signal for an area in the shape of a regular hexagon of side $$8\mbox{ km}$$, as shown in the diagram.
Find the area of the hexagon.
Give your answer in $$\mbox{km}^2$$, correct to $$2$$ decimal places.

(ii) A circle which touches all vertices of the hexagon can show areas where the signal is weak. One of these areas is shaded in the diagram.
Find this shaded area.
Give your answer in $$\mbox{km}^2$$, correct to $$1$$ decimal place.

(i) $$166.28\mbox{ km}^2$$

(ii) $$5.8\mbox{ km}^2$$

Solution

(i)

\begin{align}A&=6\times\left[\frac{1}{2}(8)(8)(\sin60^{\circ})\right]\\&\approx166.28\mbox{ km}^2\end{align}

(ii)

\begin{align}A&=\frac{\pi(8^2)-166.28}{6}\\&\approx5.8\mbox{ km}^2\end{align}

Video Walkthrough

## 2019 Paper 2 Question 9

$$R$$ is a radar station located $$120\mbox{ km}$$ north of a port $$P$$.
The circle $$c$$, centred at $$R$$ and with radius $$100\mbox{ km}$$ shows the detection range of the radar. When a ship enters the circle it will be detected by the radar station at $$R$$.
Figure $$1$$ shows a ship leaving port $$P$$ and sailing in the direction north $$30^{\circ}$$ east.
The ship enters the circle $$c$$ at $$S$$ and exits at $$T$$.
At $$Q$$, the ship is closest to $$R$$ and $$|\angle PQR=90^{\circ}$$.

(a) The triangle $$PQR$$ taken from Figure 1 is shown in Figure 2.
Find $$|QR|$$, the length of $$[QR]$$.

$$60\mbox{ km}$$

Solution

\begin{align}\sin30^{\circ}=\frac{|QR|}{120}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|QR|&=120\sin30^{\circ}\\&=60\mbox{ km}\end{align}

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(b) The triangle $$QRS$$ taken from Figure 1 is shown in Figure 3.
Use your answer from part (a) to find $$|QS|$$.

$$80\mbox{ km}$$

Solution

\begin{align}|QS|&=\sqrt{|RS|^2-|RQ|^2}\\&=\sqrt{100^2-60^2}\\&=80\mbox{ km}\end{align}

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(c) Find $$|PS|$$. Give your answer correct to the nearest $$\mbox{km}$$.

$$24\mbox{ km}$$

Solution

\begin{align}|PS|&=|PQ|-|QS|\\&=\sqrt{|PR|^2-|QR|^2}-|QS|\\&=\sqrt{120^2-60^2}-80\\&\approx24\mbox{ km}\end{align}

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(d)

(i) Consider the triangle $$RST$$.
Use the Cosine Rule to find an expression for $$\cos \theta$$,
where $$\theta$$ is the measure of the angle $$TRS$$.
Hence show that $$\theta=106^{\circ}$$, correct to the nearest degree.

(ii) John sails directly from $$S$$ to $$T$$. Mary sails from $$S$$ to $$T$$ along the minor arc $$ST$$.
Find the difference between the distance that John sails and the distance that Mary sails. Give your answer correct to the nearest $$\mbox{ km}$$.

(iii) The sea in this region is estimated to have an average of $$1$$ ship per $$25$$ square kilometres at any time.
Use this estimate to find the number of ships in the sector $$RST$$.
Give your answer correct to the nearest whole number.

(i) $$\cos\theta=-\dfrac{7}{25}$$

(ii) $$25\mbox{ km}$$

(iii) $$370$$

Solution

(i)

\begin{align}|TS|&=2|QS|\\&=2(80)\\&=160\mbox{ km}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}160^2=100^2+100^2-2(100)(100)\cos\theta\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos\theta&=\frac{160^2-100^2-100^2}{2(100)(100)}\\&=-\frac{7}{25}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=\cos^{-1}\left(\frac{7}{25}\right)\\&\approx106^{\circ}\end{align}

as required.

(ii)

\begin{align}2\pi(100)\times\frac{106^{\circ}}{360^{\circ}}-160\approx25\mbox{ km}\end{align}

(iii)

\begin{align}\mbox{Area}&=\pi(100)^2\times\frac{106^{\circ}}{360^{\circ}}\\&=9{,}250.245…\mbox{ km}^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}N&=\frac{9{,}250.245…}{25}\\&\approx370\end{align}

Video Walkthrough

## 2018 Paper 2 Question 8(a)-(d)

The diagram shows a section of a garden divided into three parts.
In the diagram: $$|Pr|=3.3\mbox{ m}$$, $$|PQ|=6.5\mbox{ m}$$, $$|QT|=|QS|=8\mbox{ m}$$, $$|\angle QRP|=90^{\circ}$$, $$|\angle PQR|=\alpha^{\circ}$$, and $$|\angle RQS|=\beta^{\circ}$$.

(a) Use the theorem of Pythagoras to find $$|RQ|$$.

$$5.6\mbox{ m}$$

Solution

\begin{align}|RQ|&=\sqrt{|PQ|^2-|PR|^2}\\&=\sqrt{6.5^2-3.3^2}\\&=5.6\mbox{ m}\end{align}

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(b) Show that $$\alpha=31^{\circ}$$, correct to the nearest degree.

The answer is already in the question!

Solution

\begin{align}\alpha&=\sin^{-1}\left(\frac{3.3}{6.5}\right)\\&\approx31^{\circ}\end{align}

as required.

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(c) Use the value of $$\alpha$$ given in part (b) to find the value of $$\beta$$.

$$149^{\circ}$$

Solution

\begin{align}\beta&=180^{\circ}-31^{\circ}\\&=149^{\circ}\end{align}

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(d) Use the Cosine Rule to find the length of $$[RS]$$.
Give your answer correct to the nearest metre.

$$13\mbox{ m}$$

Solution

\begin{align}|RS|&=\sqrt{|RQ|^2+|QS|^2-2|RQ||QS|\cos\beta}\\&=\sqrt{5.6^2+8^2-2(5.6)(8)\cos(149^{\circ})}\\&\approx13\mbox{ m}\end{align}

Video Walkthrough

## 2017 Paper 2 Question 3(c)

The points $$A(2,1)$$, $$B(6,3)$$, $$C(5,5)$$, and $$D(1,3)$$ are the vertices of the rectangle $$ABCD$$ as shown.

(c) Use trigonometry to find the measure of the angle $$ABD$$.
Give your answer correct to the nearest degree.

$$27^{\circ}$$

Solution

\begin{align}\downarrow\end{align}

\begin{align}|\angle ABD|&=\tan^{-1}\left(\frac{1}{2}\right)\\&\approx27^{\circ}\end{align}

Video Walkthrough

## 2017 Paper 2 Question 6

(a) Find the distance $$x$$ in the diagram below (not to scale).
Give your answer correct to $$2$$ decimal places.

$$8.84\mbox{ cm}$$

Solution

\begin{align}\frac{x}{\sin(180^{\circ}-65^{\circ}-63^{\circ})}=\frac{10}{\sin63^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{10\sin52^{\circ}}{\sin63^{\circ}}\\&\approx8.84\mbox{ cm}\end{align}

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(b) Find the distance $$y$$ in the diagram below (not to scale).
Give your answer correct to $$2$$ decimal places.

$$8.60\mbox{ cm}$$

Solution

\begin{align}y&=\sqrt{8.5^2+10.2^2-2(8.5)(10.2)\cos53.8^{\circ}}\\&\approx8.60\mbox{ cm}\end{align}

Video Walkthrough

## 2016 Paper 2 Question 2

(a) Find the area of the given triangle.

$$24\mbox{ cm}^2$$

Solution

\begin{align}A&=\frac{1}{2}ab\sin C\\&=\frac{1}{2}(8)(12)\sin30^{\circ}\\&=24\mbox{ cm}^2\end{align}

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(b) A triangle has sides of length $$3\mbox{ cm}$$, $$5\mbox{ cm}$$, and $$7\mbox{ cm}$$.
Find the size of the largest angle in the triangle.

$$120^{\circ}$$

Solution

\begin{align}7^2=5^2+3^2-2(5)(3)\cos A\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A&=\cos^{-1}\left(\frac{5^2+3^2-7^2}{2(5)(3)}\right)\\&=120^{\circ}\end{align}

Video Walkthrough

## 2016 Paper 2 Question 6(c)

(c) The lengths of the sides of a right-angled triangle are $$5$$, $$x$$ and $$x+1$$ as shown.
Use the Theorem of Pythagoras to find the value of $$x$$.

$$x=12$$

Solution

\begin{align}(x+1)^2=5^2+x^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2+2x+1=25+x^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x=24\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=12\end{align}

Video Walkthrough

## 2016 Paper 2 Question 9(e)

Various landmarks of Joe’s farm are marked on the diagram below.

(e) Given that $$|\angle QFB|=45^{\circ}$$, $$|BF|=5$$ units and $$|QB|=6.08$$ units, use trigonometric methods to find $$|\angle BQF|$$.
Give your answer in degrees correct to one decimal place.

$$35.6^{\circ}$$

Solution

\begin{align}\frac{\sin|\angle BQF|}{5}=\frac{\sin45^{\circ}}{6.08}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|\angle BQF|&=\sin^{-1}\left(\frac{5\sin45^{\circ}}{6.08}\right)\\&\approx35.6^{\circ}\end{align}

Video Walkthrough

## 2015 Paper 2 Question 5

The diagram shows the triangles $$BCD$$ and $$ABD$$, with some measurements given.

(a)

(i) Find $$|BC|$$, correct to two
decimal places.

(ii) Find the area of the triangle $$BCD$$, correct to two decimal places.

(i) $$11.39\mbox{ m}$$

(ii) $$42.78\mbox{ m}^2$$

Solution

(i)

\begin{align}\frac{|BC|}{\sin42^{\circ}}=\frac{16}{\sin110^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|BC|&=\frac{16\sin42^{\circ}}{\sin110^{\circ}}\\&\approx11.39\mbox{ m}\end{align}

(ii)

\begin{align}A&=ab\sin C\\&=(11.39)(16)\sin(180^{\circ}-110^{\circ}-42^{\circ})\\&\approx42.78\mbox{ m}^2\end{align}

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(b) Find $$|AB|$$, correct to two decimal places.

$$16.53\mbox{ m}$$