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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) $$60^{\circ}$$

(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.

(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 $$C$$.
Remember that $$|BC|=4.65\mbox{ m}$$, correct to $$2$$ decimal places.

(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 $$ABCDE$$.

$$\mbox{Area}=10\mbox{ cm}^2$$ 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.

$$118.74^{\circ}$$

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 $$E$$.

(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 $$y$$.

(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.

(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.
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 $$RST$$.

(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.

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 $$[RS]$$.

$$13\mbox{ m}$$

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 $$ABD$$.

$$27^{\circ}$$

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 $$|\angle BQF|$$.

$$35.6^{\circ}$$

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.

$$16.53\mbox{ m}$$