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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... |
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Area of a Triangle |
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Sine/Cosine Rules |
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3D Problems |
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Trigonometric Equations |
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Trigonometric Identities |
Has not appeared |
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Compound Angle Formulae |
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Double-Angle Formulae |
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Sum/Difference/Product Formulae |
Has not appeared |
2022 Paper 1 Question 8(e)
A Ferris wheel has a diameter of \(120\mbox{ m}\).
When it is turning, it completes exactly \(10\) full rotations in one hour.
The diagram above shows the Ferris wheel before it starts to turn.
At this stage, the point \(A\) is the lowest point on the circumference of the wheel, and it is at a
height of \(12\mbox{ m}\) above ground level.
The height, \(h\), of the point \(A\) after the wheel has been turning for \(t\) minutes is given by:
\begin{align}h(t)=72-60\cos\left(\frac{\pi}{3}t\right)\end{align}
where \(h\) is in metres, \(t\in\mathbb{R}\), and \(\dfrac{\pi}{3}\) is in radians.
(e) By solving the following equation, find the second time (value of \(t\)) that the point \(A\) is at a height of \(110\mbox{ m}\), after it starts turning:
\begin{align}72-60\cos\left(\frac{\pi}{3}t\right)=110\end{align}
Give your answer in minutes, correct to \(2\) decimal places.
\(3.85\mbox{ minutes}\)
\begin{align}72-60\cos\left(\frac{\pi}{3}t\right)=110\end{align}
\begin{align}\downarrow\end{align}
\begin{align}60\cos\left(\frac{\pi}{3}t\right)=-38\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos\left(\frac{\pi}{3}t\right)=-\frac{38}{60}\end{align}
\[\,\]
Reference Angle
\begin{align}\cos\left(A\right)=\frac{38}{60}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\cos^{-1}\left(\frac{38}{60}\right)\\&=0.8849…\mbox{ rad}\end{align}
Cosine is negative in 2nd quadrant (1st time) and 3rd quadrant (2nd time).
\[\,\]
Third Quadrant
\begin{align}\frac{\pi}{3}t=\pi+0.8849…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{3(\pi+0.8849…)}{\pi}\\&\approx3.85\mbox{ min}\end{align}
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2022 Paper 2 Question 4
(a)
(i) Prove that \(\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A \tan B}\).
(ii) Write \(\tan 15^{\circ}\) in the form \(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\), where \(a\in\mathbb{N}\).
(i) The answer is already in the question!
(ii) \(\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
(i)
\begin{align}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\end{align}
Replacing \(B\) with \(-B\):
\begin{align}\tan(A+(-B))=\frac{\tan A+\tan (-B)}{1-\tan A\tan (-B)}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\end{align}
where we have used \(\tan(-B)=-\tan B\).
(ii)
\begin{align}\tan15^{\circ}&=\tan(60^{\circ}-45^{\circ})\\&=\frac{\tan 60^{\circ}-\tan 45^{\circ}}{1+(\tan 60^{\circ})(\tan 45^{\circ})}\\&=\frac{\sqrt{3}-1}{1+(\sqrt{3})(1)}\\&=\frac{\sqrt{3}-1}{\sqrt{3}+1}\end{align}
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(b) The triangle \(ABC\) is shown in the diagram below.
\(|AC|=|BC|\) and \(|\angle ACB|=45^{\circ}\).
\(|AB|=10\sqrt{2-\sqrt{2}}\), as shown.
Find the length \(|AC|\).
\(10\)
As the triangle is isosceles, the other two angles are
\begin{align}\frac{180^{\circ}-45^{\circ}}{2}=67.5^{\circ}\end{align}
\[\,\]
Sine Rule
\begin{align}\frac{|AC|}{\sin67.5^{\circ}}=\frac{10\sqrt{2-\sqrt{2}}}{\sin45^{\circ}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|AC|&=\frac{10\sqrt{2-\sqrt{2}}(\sin67.5^{\circ})}{\sin45^{\circ}}\\&=10\end{align}
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2022 Paper 2 Question 9
Oscar is taking some measurements and is using trigonometry to work out some angles, distances, and areas.
First, Oscar takes measurements of two adjacent triangular fields, Field 1 (\(ABC\)) and Field 2 (\(BDC\)), as shown in the diagram below (not to scale).
\(B\) lies on the line \(AD\).
\(|AB|=30\mbox{ m}\).
\(|BD|=10\mbox{ m}\).
\(|AC|=35\mbox{ m}\).
\(|\angle CAD|=50^{\circ}\).
Note: the angle \(ABC\) is not a right angle.
(a) Find the area of Field 1 and, hence, find the area of Field 2.
Give each answer correct to the nearest \(\mbox{m}^2\).
\(A_1=402\mbox{ m}^2\) and \(A_2=134\mbox{ m}^2\)
\begin{align}A_1&=\frac{1}{2}|AC||AB|\sin A\\&=\frac{1}{2}(35)(30)(\sin 50^{\circ})\\&=402.1…\\&\approx402\mbox{ m}^2\end{align}
and
\begin{align}A_{\mbox{total}}&=\frac{1}{2}|AC||AD|\sin A\\&=\frac{1}{2}(35)(30+10)(\sin 50^{\circ})\\&=536.2…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A_2&=A_{\mbox{total}}-A_1\\&=536.2…-402.1…\\&\approx134\mbox{ m}^2\end{align}
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(b) Find the length of the perimeter of Field 1.
Give your answer correct to the nearest metre
\(93\mbox{ m}\)
\begin{align}|CB|&=\sqrt{|CA|^2+|AB|^2-2|CA|AB|\cos A}\\&=\sqrt{35^2+30^2-2(35)(3)(\cos50^{\circ})}\\&\approx 28\mbox{ m}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\mbox{Perimeter}&=35+30+28\\&=93\mbox{ m}\end{align}
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Oscar is also watching an airplane, P, fly directly over his head. He is standing at the point O in the diagrams below. The \(x\)-axis is the horizontal ground and the \(y\)-axis runs vertically up from Oscar.
The airplane P is flying at a constant height of \(10\mbox{ km}\) above the ground, as shown in the diagrams.
(c) Sound travels at a speed of roughly \(343\) metres per second in air.
(i) As the airplane flies, its engine makes noise. It takes some time for this sound to reach Oscar. Use the information in Diagram 1 to show that it takes \(41\) seconds for the sound the airplane makes at \(P_1\) to reach Oscar, correct to the nearest second.
(ii) The airplane P is flying at a constant speed of \(255\) metres per second. By the time Oscar hears the sound the airplane made at the point \(P_1\) , the airplane has flown on to the point \(P_2\), as shown in Diagram 2 above.
Work out the size of the angle marked \(\theta\) in Diagram 2, correct to \(1\) decimal place.
(i) The answer is already in the question!
(ii) \(2.6^{\circ}\)
(i)
\begin{align}\cos 45^{\circ}&=\frac{10}{|P_1O|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|P_1O|&=\frac{10}{\cos 45^{\circ}}\\&=14.142…\mbox{ km}\\&\approx14{,}142\mbox{ m}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{d}{v}\\&=\frac{14{,}142}{343}\\&\approx 41\mbox{ s}\end{align}
(ii)
\begin{align}|P_1P_2|&=vt\\&=(255)(41)\\&=10{,}455\mbox{ m}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta&=\tan^{-1}\left(\frac{10{,}455-10{,}000}{10{,}000}\right)\\&\approx2.6^{\circ}\end{align}
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(d) \(P_3\) and \(P_4\) are two other points on the flightpath of airplane P.
By the time Oscar hears the sound the airplane made at the point \(P_3\), the airplane has flown on to the point \(P_4\), as shown in the diagram below (not to scale).
\(P_3\) and \(P_4\) are both a distance of \(d \mbox{ km}\) from the \(y\)-axis.
(i) Explain briefly why the following equation holds:
\begin{align}\frac{\sqrt{100+d^2}}{0.343}=\frac{2d}{0.255}\end{align}
(ii) Solve the equation above to find the value of \(d\), correct to \(1\) decimal place.
(i) The left hand side is the time taken for the sound to go from \(P_3\) to \(O\). The right hand side is the time taken for the plane to go from from \(P_3\) to \(P_4\). These should be the same.
(ii) \(4.0\mbox{ km}\)
(i) The left hand side is the time taken for the sound to go from \(P_3\) to \(O\). The right hand side is the time taken for the plane to go from from \(P_3\) to \(P_4\). These should be the same.
(ii)
\begin{align}\frac{\sqrt{100+d^2}}{0.343}=\frac{2d}{0.255}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sqrt{100+d^2}&=\frac{2d(0.343)}{0.255}\\&=(2.69…)d\end{align}
\begin{align}\downarrow\end{align}
\begin{align}100+d^2=(7.23…)d^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(6.23…)d^2=100\end{align}
\begin{align}\downarrow\end{align}
\begin{align}d&=\sqrt{\frac{100}{6.23…}}\\&\approx4.0\mbox{ km}\end{align}
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2021 Paper 2 Question 4
(a)
(i) Prove that \(\cos 2A=\cos^2A-\sin^2A\).
(ii) \(\sin\left(\dfrac{\theta}{2}\right)=\dfrac{1}{\sqrt{5}}\), where \(0\leq\theta\leq\pi\).
Use the formula \(\cos 2A=\cos^2A-\sin^2A\) to find the value of \(\cos\theta\).
(i) The answer is already in the question!
(ii) \(\cos\theta=\dfrac{3}{5}\)
(i)
\begin{align}\cos(A+B)&=\cos A\cos B-\sin A\sin B\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos(A+A)&=\cos A\cos A-\sin A\sin A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos(2A)&=\cos^2A-\sin^2A\end{align}
as required.
(ii)
\begin{align}\cos(2A)&=\cos^2A-\sin^2A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos(\theta)&=\cos^2\left(\frac{\theta}{2}\right)-\sin^2\left(\frac{\theta}{2}\right)\\&=\left[1-\sin^2\left(\frac{\theta}{2}\right)\right]-\sin^2\left(\frac{\theta}{2}\right)\\&=1-2\sin^2\left(\frac{\theta}{2}\right)\\&=1-2\left(\frac{1}{\sqrt{5}}\right)^2\\&=1-\frac{2}{5}\\&=\frac{3}{5}\end{align}
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(b) Solve the equation:
\begin{align}\tan(B+150^{\circ})=-\sqrt{3}\end{align}
for \(0^{\circ}\leq B\leq360^{\circ}\).
\(B=150^{\circ}\) and \(B=330^{\circ}\)
\begin{align}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan(B+150^{\circ})=\frac{\tan B+\tan 150^{\circ}}{1-\tan B\tan 150^{\circ}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{\tan B+\tan 150^{\circ}}{1-\tan B\tan 150^{\circ}}=-\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{\tan B+\left(-\frac{1}{\sqrt{3}}\right)}{1-\tan B\left(-\frac{1}{\sqrt{3}}\right)}=-\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan B-\frac{1}{\sqrt{3}}=-\sqrt{3}-\tan B\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2\tan B&=-\sqrt{3}+\frac{1}{\sqrt{3}}\\&=\frac{-3+1}{\sqrt{3}}\\&=-\frac{2}{\sqrt{3}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan B=-\frac{1}{\sqrt{3}}\end{align}
For the reference angle \(0^{\circ}\leq A\leq 90^{\circ}\):
\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A=30^{\circ}\end{align}
Tangent is negative in both the second and fourth quadrants.
\[\,\]
Second Quadrant
\begin{align}B&=180^{\circ}-30^{\circ}\\&=150^{\circ}\end{align}
\[\,\]
Fourth Quadrant
\begin{align}B&=360^{\circ}-30^{\circ}\\&=330^{\circ}\end{align}
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2021 Paper 2 Question 7(c), (d) & (f)
The diagram (Triangle \(ABC\)) shows the \(3\) sections of a level triathlon course.
In order to complete the triathlon, each contestant must swim \(4\mbox{ km}\) from \(C\) to \(B\), cycle from \(B\) to \(A\), and then run \(28\mbox{ km}\) from \(A\) to \(C\).
Mary can cycle at an average speed of \(25\mbox{ km/hour}\).
It takes her \(1\) hour and \(12\) minutes to cycle from \(B\) to \(A\).
(c) Show that \(|\angle ACB=116.5^{\circ}|\), correct to \(1\) decimal place.
The answer is already in the question!
\begin{align}|AB|^2=|AC|^2+|BC|^2-2|AC||BC|\cos|\angle ACB|\end{align}
\begin{align}\downarrow\end{align}
\begin{align}30^2=28^2+4^2-2(28)(4)\cos|\angle ACB|\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|\angle ACB|&=\cos^{-1}\left(\frac{28^2+4^2-30^2}{2(28)(4)}\right)\\&\approx116.5^{\circ}\end{align}
as required.
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(d) To comply with safety regulations, the region inside the triangular course must be kept clear of people. Find the area of this region.
Give your answer, in \(\mbox{km}^2\), correct to \(1\) decimal place.
\(50.1\mbox{ km}^2\)
\begin{align}\mbox{Area}&=\frac{1}{2}|AC||BC|\sin |\angle ACB|\\&=\frac{1}{2}(28)(4)\sin116.5{^\circ}\\&\approx50.1\mbox{ km}^2\end{align}
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(f) The course is viewed from a camera tower which rises vertically from point \(A\).
The top of the tower is point \(T\). The angle of elevation of \(T\) from \(B\) is \(0.05^{\circ}\).
Find \(|AT|\), the vertical height of the tower.
Give your answer correct to the nearest metre.
\(26\mbox{ m}\)
\begin{align}\tan(0.05^{\circ})&=\frac{|AT|}{|AB|}\\&=\frac{|AT|}{30}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|AT|&=30\tan(0.05^{\circ})\\&=0.02617…\mbox{ km}\\&\approx 26\mbox{ m}\end{align}
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2021 Paper 2 Question 9
(a) An aeroplane flies east from point \(A\) for \(2\) hours at a constant speed of \(420\) km per hour until it reaches point \(B\). It then changes direction by heading \(20^{\circ}\) towards the south at the same speed until it reaches point \(C\), as shown in the diagram below.
The direct distance from \(A\) to \(C\) is \(1450\mbox{ km}\) and \(|\angle BAC|=8.57^{\circ}\).
(i) Find how long it took to fly from \(B\) to \(C\).
Give your answer correct to the nearest minute.
(ii) The average fuel consumption of the plane is \(3.8\) litres per second and the fuel capacity of the plane is \(100{,}000\) litres.
Show that the plane will be able to complete the journey from \(A\) to \(B\) to \(C\) and directly back to \(A\) at a speed of \(420\mbox{ km/h}\) without refuelling.
(i) \(1\mbox{ h }30\mbox{ min}\)
(ii) The answer is already in the question!
(i)
\begin{align}|AB|&=(2)(420)\\&=840\mbox{ km}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|BC|^2=|AB|^2+|AC|^2-2|AB||AC|\cos |\angle BAC|\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|BC|&=\sqrt{840^2+1450^2-2(840)(1450)(\cos8.57^{\circ}}\\&=631.90…\mbox{ km}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{631.90…}{420}\\&=1.50…\mbox{ h}\\&\approx1\mbox{ h }30\mbox{ min}\end{align}
(ii)
\begin{align}T&=t_{AB}+t_{BC}+t_{CA}\\&=2+1.5+\frac{1450}{420}\\&=6.9523…\mbox{ h}\\&=25{,}028.57…\mbox{ s}\end{align}
As the fuel runs out in a longer time of \(\dfrac{100{,}000}{3.8}=26{,}315.7…\mbox{ s}\), the plane can complete the journey.
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(b) The voltage, \(V(t)\), (in Volts) of a certain alternating current is given by the function:
\begin{align}V(t)=110\sqrt{2}\sin(120\pi t)\end{align}
where \(t\) is in seconds.
(i) Find the period and range of the function \(V(t)\).
(ii) Sketch the function for \(0\leq t \leq p\), where \(p\) is the period of \(V(t)\).
Indicate the period and range of the function on your graph.
(iii) Use \(V(t)\) to find the voltage when \(t=6.67\) seconds.
Give your answer correct to two decimal places.
(iv) Find one value for \(t\) where the voltage is \(110\) Volts.
Give your answer in the form \(\dfrac{a}{b}\) where \(a,b\in\mathbb{N}\).
(v) Find the rate of change of the voltage when \(t=2\) seconds.
Give your answer correct to the nearest unit.
(i) The period is \(\dfrac{1}{60}\mbox{ s}\) and the range is \([-110\sqrt{2},110\sqrt{2}]\).
(ii)
(iii) \(147.95\mbox{ Volts}\)
(iv) \(t=\dfrac{1}{480}\mbox{ s}\)
(v) \(58{,}646\mbox{ Volts/sec}\)
(i)
\begin{align}\mbox{Period}&=\frac{2\pi}{\omega}\\&=\frac{2\pi}{120\pi}\\&=\frac{1}{60}\mbox{ s}\end{align}
and the range is \([-110\sqrt{2},110\sqrt{2}]\).
(ii)
(iii)
\begin{align}V(6.67)&=110\sqrt{2}\sin[120\pi(6.67)]\\&\approx147.95\mbox{ Volts}\end{align}
(iv)
\begin{align}110\sqrt{2}\sin(120\pi t)=110\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sin(120\pi t)=\frac{1}{\sqrt{2}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}120\pi t=\frac{\pi}{4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{1}{480}\mbox{ s}\end{align}
(v)
\begin{align}V'(t)=(120\pi)[110\sqrt{2}\cos(120\pi t)]\end{align}
\begin{align}\downarrow\end{align}
\begin{align}V'(2)&=(120\pi)[110\sqrt{2}\cos(120\pi (2))]\\&\approx 58{,}646\mbox{ Volts/sec}\end{align}
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2020 Paper 1 Question 8(a)
A rectangle is inscribed in a circle of radius \(5\) units and centre \(O(0,0)\) as shown below.
Let \(R(x,y)\), where \(x,y\in\mathbb{R}\), be the vertex of the rectangle in the first quadrant as shown.
Let \(\theta\) be the angle between \([OR]\) and the positive \(x\)-axis, where \(0\leq\theta\leq\dfrac{\pi}{2}\).
(a)
(i) The point \(R(x,y)\) can be written as \(a\cos \theta, b\sin\theta\), where \(a,b\in\mathbb{R}\).
Find the value of \(a\) and the value of \(b\).
(ii) Show that\(A(\theta)\), the area of the rectangle, measured in square units, can be written as \(A(\theta)=50\sin2\theta\).
(iii) Use calculus to show that the rectangle with maximum area is a square.
(iv) Find this maximum area.
(i) \(a=5\) and \(b=5\)
(ii) The answer is already in the question!
(iii) The answer is already in the question!
(iv) \(50\mbox{ square units}\)
(i)
\begin{align}x=5\cos\theta&&y=5\sin\theta\end{align}
\begin{align}\downarrow\end{align}
\(a=5\) and \(b=5\)
(ii)
\begin{align}A(\theta)&=(2\times5\sin\theta)(2\times5\cos\theta)\\&=100\sin\theta\cos\theta\\&=50(2\sin\theta\cos\theta)\\&=50\sin2\theta\end{align}
(iii)
Setting the derivative to zero, we obtain:
\begin{align}(50\cos2\theta)(2)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos2\theta=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2\theta&=\cos^{-1}(0)\\&=90^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta=45^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}l&=10\cos45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}
and
\begin{align}w&=10\sin45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}
As these are the same, the maximum area is a square.
(iv)
\begin{align}A&=l\times w\\&=\left(\frac{10}{\sqrt{2}}\right)\times\left(\frac{10}{\sqrt{2}}\right)\\&=50\mbox{ square units}\end{align}
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2020 Paper 2 Question 4
(a) Find the two values of \(\theta\) for which \(\tan \theta=-\dfrac{1}{\sqrt{3}}\), where \(0\leq \theta \leq 4\pi\).
\(\dfrac{5\pi}{3}\) and \(\dfrac{11\pi}{3}\)
Let \(\alpha=\dfrac{\theta}{2}\).
\begin{align}\tan \alpha=-\frac{1}{\sqrt{3}}\end{align}
\[\,\]
Reference Angle
\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A=\frac{\pi}{6}\end{align}
Tangent is negative in both the second and fourth quadrants.
\[\,\]
Second Quadrant
\begin{align}\alpha&=\pi-\frac{\pi}{6}\\&=\frac{5\pi}{6}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{\theta}{2}=\frac{5\pi}{6}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta=\frac{5\pi}{3}\end{align}
\[\,\]
Fourth Quadrant
\begin{align}\alpha&=2\pi-\frac{\pi}{6}\\&=\frac{11\pi}{6}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{\theta}{2}=\frac{11\pi}{6}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta=\frac{11\pi}{3}\end{align}
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(b) The diagram shows \(OAB\), a sector of a circle of radius \(7\mbox{ cm}\) with centre \(O\).
In the sector, \(|\angle BOA|=1.2\) radians.
The area of the shaded region is \(21\mbox{ cm}^2\).
Find \(|BC|\).
Give your answer correct to \(1\) decimal place.
\(4.4\mbox{ cm}\)
\begin{align}A_{ABO}=A_{ABC}+A_{ACO}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}r^2\theta=21+\frac{1}{2}|CO||AO|\sin(|\angle AOC|)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}(7^2)(1.2)=21+\frac{1}{2}|CO|(7)\sin(1.2\mbox{ rad})\end{align}
\begin{align}\downarrow\end{align}
\begin{align}29.4=21+3.5|CO|\sin(1.2\mbox{ rad})\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|CO|&=\frac{29.4-21}{3.5\sin(1.2\mbox{ rad})}\\&=2.57…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|BC|&=|BO|-|CO|\\&=7-2.57…\\&\approx4.4\mbox{ cm}\end{align}
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2019 Paper 2 Question 4
(a) Show that \(\cos2\theta=1-2\sin^2\theta\).
The answer is already in the question!
\begin{align}\cos(A+B)=\cos A\cos B-\sin A\sin B\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos(\theta+\theta)=\cos\theta\cos\theta-\sin\theta\sin\theta\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos2\theta&=\cos^2\theta-\sin^2\theta\\&=(1-\sin^2\theta)-\sin^2\theta\\&=1-2\sin^2\theta\end{align}
as required.
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(b) Find the cosine of the acute angle between two diagonals of a cube.
\(\dfrac{1}{3}\)
Let \(l\) be the length of each edge.
The diagonal \(d\) on each face is then
\begin{align}d&=\sqrt{x^2+x^2}\\&=\sqrt{2}x\end{align}
The two internal diagonals \(D\) are therefore
\begin{align}D&=\sqrt{x^2+(\sqrt{2}x)^2}\\&=\sqrt{x^2+2x^2}\\&=\sqrt{3}x\end{align}
\begin{align}\downarrow\end{align}
\[\,\]
Cosine Rule
\begin{align}x^2=\left(\frac{\sqrt{3}x}{2}\right)^2+\left(\frac{\sqrt{3}x}{2}\right)^2-2\left(\frac{\sqrt{3}x}{2}\right)\left(\frac{\sqrt{3}x}{2}\right)\cos A\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos A&=\frac{\left(\frac{\sqrt{3}x}{2}\right)^2+\left(\frac{\sqrt{3}x}{2}\right)^2-x^2}{2\left(\frac{\sqrt{3}x}{2}\right)\left(\frac{\sqrt{3}x}{2}\right)}\\&=\frac{\frac{1}{2}}{\frac{3}{2}}\\&=\frac{1}{3}\end{align}
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2019 Paper 2 Question 9
The diagram below shows a triangular patch of ground \(\Delta SGH\), with \(|SH|=58\mbox{ m}\), \(|GH=30\mbox{ m}\),
and \(|\angle GHS|=68^{\circ}\). The circle is a helicopter pad. It is the incircle of \(\Delta SGH\) and has centre \(P\).
(a) Find \(|SG|\). Give your answer in metres, correct to1decimal place.
\(54.4\mbox{ m}\)
\begin{align}|SG|&=\sqrt{|GH|^2+|SH|^2-2|GH||SH|\cos68^{\circ}}\\&=\sqrt{30^2+58^2-2(30)(58)\cos68^{\circ}}\\&\approx54.4\mbox{ m}\end{align}
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(b) Find \(|\angle HSG|\). Give your answer in degrees, correct to \(2\) decimal places.
\(30.75^{\circ}\)
\begin{align}\frac{\sin|\angle HSG|}{30}=\frac{\sin68^{\circ}}{54.4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sin|\angle HSG|=30\left(\frac{\sin68^{\circ}}{54.4}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sin|\angle HSG|=0.51131…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|\angle HSG|&=\sin^{-1}(0.51131…)\\&\approx30.75^{\circ}\end{align}
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(c) Find the area of \(\Delta SGH\). Give your answer in \(\mbox{m}^2\), correct to \(2\) decimal places
\(806.65\mbox{ m}^2\)
\begin{align}A&=\frac{1}{2}ab\sin C\\&=\frac{1}{2}(30)(58)\sin68^{\circ}\\&\approx806.65\mbox{ m}^2\end{align}
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(d)
(i) Find the area of \(\Delta HSP\), in terms of \(r\), where \(r\) is the radius of the helicopter pad.
(ii) Show that the area of \(\Delta SGH\), in terms of \(r\), can be written as \(71.2r\mbox{ m}^2\).
(iii) Find the value of \(r\). Give your answer in metres, correct to \(1\) decimal place.
(i) \(29r\mbox{ m}^2\)
(ii) The answer is already in the question!
(iii) \(11.3\mbox{ m}\)
(i)
\begin{align}A&=\frac{1}{2}bh\\&=\frac{1}{2}(58)(r)\\&=29r\mbox{ m}^2\end{align}
(ii)
\begin{align}A&=\frac{1}{2}(58)(r)+\frac{1}{2}(30)(r)+\frac{1}{2}(54.4)(r)\\&=71.2r\mbox{ m}^2\end{align}
as required.
(iii)
\begin{align}71.2r=806.62\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r&=\frac{806.62}{71.2}\\&\approx11.3\mbox{ m}\end{align}
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(e) \([ST]\) is a vertical pole at the point \(S\).
The angle of elevation of the top of the pole from the point \(P\) is \(14^{\circ}\).
Find the height of the pole.
Give your answer, in metres, correct to \(1\) decimal place.
\(10.7\mbox{ m}\)
\begin{align}\sin\left(\frac{30.75^{\circ}}{2}\right)=\frac{11.3}{|PS|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|PS|&=\frac{11.3}{\sin\left(\frac{30.75^{\circ}}{2}\right)}\\&=42.619…\end{align}
and
\begin{align}\tan14^{\circ}=\frac{|ST|}{|PS|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|ST|&=|PS|\tan14^{\circ}\\&=(42.619…)(\tan14^{\circ})\\&\approx10.7\mbox{ m}\end{align}
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2018 Paper 2 Question 4
(a) Find all the values of \(x\) for which \(\cos(2x)=-\dfrac{\sqrt{3}}{2}\), where \(0^{\circ}\leq x\leq 360^{\circ}\).
\(x=75^{\circ}\), \(x=105^{\circ}\), \(x=255^{\circ}\) and \(x=285^{\circ}\)
Let \(\theta=2x\).
\begin{align}\cos\theta=-\frac{\sqrt{3}}{2}\end{align}
\begin{align}\downarrow\end{align}
Reference Angle:
\begin{align}\cos A=\frac{\sqrt{3}}{2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A=30^{\circ}\end{align}
\[\,\]
Second Quadrant
\begin{align}\theta&=(180^{\circ}-30^{\circ})\\&=150^{\circ}+360^{\circ}n\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{\theta}{2}\\&=\frac{150^{\circ}+360^{\circ}n}{2}\\&=75^{\circ}+180^{\circ}n\end{align}
\begin{align}\downarrow\end{align}
\(x=75^{\circ}\) and \(x=255^{\circ}\)
\[\,\]
Third Quadrant
\begin{align}\theta&=(180^{\circ}+30^{\circ})+360^{\circ}n\\&=210^{\circ}+360^{\circ}n\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{\theta}{2}\\&=\frac{210^{\circ}+360^{\circ}n}{2}\\&=105^{\circ}+180^{\circ}n\end{align}
\begin{align}\downarrow\end{align}
\(x=105^{\circ}\) and \(x=285^{\circ}\)
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(b) Let \(\cos A=\dfrac{y}{2}\), where \(0^{\circ}<A<90^{\circ}\). Write \(\sin2A\) in terms of \(y\).
\(\dfrac{y\sqrt{4-y^2}}{2}\)
\begin{align}\sin A=\frac{\sqrt{4-y^2}}{2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sin(2A)&=2\sin A\cos A\\&=2\left(\frac{\sqrt{4-y^2}}{2}\right)\left(\frac{y}{2}\right)\\&=\frac{y\sqrt{4-y^2}}{2}\end{align}
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2017 Paper 2 Question 6
(a) Take the earth as a sphere with radius \(6371\mbox{ km}\).
Jack is standing on the Cliffs of Moher at the point \(J\) which is \(214\) metres above sea level.
He is looking out to sea at a point \(H\) on the horizon.
Taking \(A\) as the centre of the earth, find \(|JH|\), the distance from Jack to the horizon.
Give your answer correct to the nearest \(\mbox{km}\).
\(52\mbox{ km}\)
As \(|\angle AHJ|\) is a right angle:
\begin{align}|JH|&=\sqrt{|AJ|^2-|AH|^2}\\&=\sqrt{(6371+0.214)^2-6371^2}\\&\approx52\mbox{ km}\end{align}
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(b) The Cliffs of Moher, at point \(C\), are at latitude \(53^{\circ}\) north of the equator.
On the diagram, \(s_1\) represents the circle that is at latitude \(53^{\circ}\).
\(s_2\) represents the equator (which is at latitude \(0^{\circ}\)).
\(A\) is the centre of the earth.
\(s_1\) and \(s_2\) are on parallel planes.
Find the length of the circle \(s_1\).
Give your answer correct to the nearest \(\mbox{km}\).
\(24{,}091\mbox{ km}\)
\begin{align}r_{s_1}&=6371\times\cos53^{\circ}\\&=3834.1635…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}l_{s_1}&=2\pi r_{s_1}\\&=2\pi(3834.1635…)\\&\approx24{,}091\mbox{ km}\end{align}
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2017 Paper 2 Question 9(a) - (e)
Conor’s property is bounded by the straight bank of a river, as shown in Figure 1 above.
\(T\) is the base of a vertical tree that is growing near the opposite bank of the river.
\(|TE|\) is the height of the tree, as shown in Figure 2 above.
From the point \(C\), which is due west of the tree, the angle of elevation of \(E\), the top of the tree, is \(60^{\circ}\).
From the point \(D\), which is \(15\mbox{ m}\) due north of \(C\), the angle of elevation of \(E\) is \(30^{\circ}\) (see Figure 2).
The land on both sides of the river is flat and at the same level.
(a) Use triangle \(ECT\), to express \(|TE|\) in the form \(\sqrt{a}|CT|\) metres, where \(a\in\mathbb{N}\).
\(|TE|=\sqrt{3}|CT|\)
\begin{align}\tan60^{\circ}&=\frac{|TE|}{|CT|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sqrt{3}&=\frac{|TE|}{|CT|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|TE|=\sqrt{3}|CT|\end{align}
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(b) Show that \(|TE|\) may also be expressed as \(\sqrt{\dfrac{225+|CT|^2}{3}}\) metres.
The answer is already in the question!
\begin{align}\tan30^{\circ}&=\frac{|TE|}{|DT|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{\sqrt{3}}&=\frac{|TE|}{|DT|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|TE|&=\frac{|DT|}{\sqrt{3}}\\&=\frac{\sqrt{15^2+|CT|^2}}{\sqrt{3}}\\&=\sqrt{\frac{225+|CT|^2}{3}}\end{align}
as required.
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(c) Hence find \(|CT|\), the distance from the base of the tree to the bank of the river at Conor’s side.
Give your answer correct to \(1\) decimal place.
\(|CT|=5.3\mbox{ m}\)
\begin{align}\sqrt{3}|CT|=\sqrt{\frac{225+|CT|^2}{3}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3|CT|^2=\frac{225+|CT|^2}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9|CT|^2=225+|CT|^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8|CT|^2=225\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|CT|&=\sqrt{\frac{225}{8}}\\&\approx5.3\mbox{ m}\end{align}
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(d) Find \(|TE|\) the height of the tree. Give your answer correct to \(1\) decimal place.
\(9.2\mbox{ m}\)
\begin{align}|TE|&=\sqrt{3}|CT|\\&=\sqrt{3}(5.3)\\&\approx9.2\mbox{ m}\end{align}
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(e) The tree falls across the river and hits the bank at Conor’s side at the point \(F\). Find the maximum size of the angle \(FTC\). Give your answer in degrees, correct to \(1\) decimal place.
\(54.7^{\circ}\)
\begin{align}\cos A&=\frac{|CT|}{|FT|}\\&=\frac{|CT|}{|TW|}\\&=\frac{|CT|}{\sqrt{3}|CT|}\\&=\frac{1}{\sqrt{3}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\\&\approx54.7^{\circ}\end{align}
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2016 Paper 1 Question 5(a)
(a)
(i) The lengths of the sides of a right-angled triangle are given by the expressions \(x-1\), \(4x\), and \(5x-9\), as shown in the diagram.
Find the value of \(x\).
(ii) Verify, with this value of \(x\), that the lengths of the sides of the triangle above form a pythagorean triple.
(i) \(x=10\)
(ii) The answer is already in the question!
(i)
\begin{align}(x-1)^2+(4x)^2=(5x-9)^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-2x+1+16x^2=25x^2-90x+81\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8x^2-88x+80=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-11x+10=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-1)(x-10)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=1\) or \(x=10\)
\begin{align}\downarrow\end{align}
\(x=10\)
(as \(5(1)-9<0\)).
(ii)
\begin{align}(x-1)^2+(4x)^2=(5x-9)^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(10-1)^2+(4(10))^2=(5(10)-9)^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9^2+40^2=41^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1681=1681\end{align}
as required.
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2016 Paper 2 Question 3
(a) Show that \(\dfrac{\cos 7A+\cos A}{\sin 7A-\sin A}=\cot 3A\).
The answer is already in the question!
\begin{align}\frac{\cos7A+\cos A}{\sin7A-\sin A}&=\frac{\cos(4A+3A)+\cos(4A-3A)}{\sin(4A+3A)-\sin(4A-3A)}\\&=\frac{2\cos4A\cos3A}{2\cos4A\sin3A}\\&=\frac{\cos3A}{\sin3A}\\&=\cot3A\end{align}
as required.
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(b) Given that \(\cos2\theta=\dfrac{1}{9}\), find \(\cos\theta\) in the form \(\pm\dfrac{\sqrt{a}}{b}\), where \(a,b\in\mathbb{N}\).
\(\pm\dfrac{\sqrt{5}}{3}\)
\begin{align}\cos^2\theta=\frac{1}{2}(1+\cos2\theta)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos\theta&=\pm\sqrt{\frac{1}{2}(1+\cos2\theta)}\\&=\pm\sqrt{\frac{1}{2}\left(1+\frac{1}{9}\right)}\\&=\pm\sqrt{\frac{5}{9}}\\&=\pm\frac{\sqrt{5}}{3}\end{align}
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2016 Paper 2 Question 7
A glass Roof Lantern in the shape of a pyramid has a rectangular base \(CDEF\) and its apex is at \(B\) as
shown. The vertical height of the pyramid is \(|AB|\), where \(A\) is the point of intersection of the
diagonals of the base as shown in the diagram.
Also \(|CD|=2.5\mbox{ m}\) and \(|CF|=3\mbox{ m}\).
(a)
(i) Show that \(|AC|=1.95\mbox{ m}\), correct to two decimal places.
(ii) The angle of elevation of \(B\) from \(C\) is \(50^{\circ}\) (i.e. \(|\angle BCA|=50^{\circ}\)).
Show that \(|AB|=2.3\mbox{ m}\), correct to one decimal place.
(iii) Find \(|BC|\), correct to the nearest metre.
(iv) Find \(|\angle BCD|\), correct to the nearest degree.
(v) Find the area of glass required to glaze all four triangular sides of the pyramid.
Give your answer correct to the nearest \(\mbox{m}^2\).
(i) The answer is already in the question!
(ii) The answer is already in the question!
(iii) \(3\mbox{ m}\)
(iv) \(65^{\circ}\)
(v) \(15\mbox{ m}^2\)
(i)
\begin{align}|EC|^2=|ED|^2+|CD|^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(2|AC|)^2=|ED|^2+|CD|^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|AC|&=\frac{\sqrt{|ED|^2+|CD|^2}}{2}\\&=\frac{\sqrt{3^2+2.5^2}}{2}\\&\approx1.95\mbox{ m}\end{align}
(ii)
\begin{align}\tan50^{\circ}=\frac{|AB|}{1.95}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|AB|&=1.95\tan50^{\circ}\\&\approx2.3\mbox{ m}\end{align}
as required.
(iii)
\begin{align}|BC|&=\sqrt{1.95^2+2.3^2}\\&\approx3\mbox{ m}\end{align}
(iv)
\begin{align}|BD|^2=|BC|^2+|CD|^2-2|BC||CD|\cos|\angle BCD|\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|\angle BCD|&=\cos^{-1}\left(\frac{|BC|^2+|CD|^2-|BD|^2}{2|BC||CD|}\right)\\&=\cos^{-1}\left(\frac{3^2+2.5^2-3^2}{2(3)(2.5)}\right)\\&\approx65^{\circ}\end{align}
(v)
\begin{align}A&=2\times\left(\frac{1}{2}(3)(3)\sin60^{\circ}\right)+2\times\left(\frac{1}{2}(2.5)(3)\sin65^{\circ}\right)\\&\approx15\mbox{ m}^2\end{align}
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(b) Another Roof Lantern, in the shape of a pyramid, has a square base \(CDEF\). The vertical height \(|AB|=3\mbox{ m}\), where
\(A\) is the point of intersection of the diagonals of the base as shown.
The angle of elevation of \(B\) to \(C\) is \(60^{\circ}\)
(i.e. \(|\angle BCA|=60^{\circ}\)).
Find the length of the side of the square base of the lantern.
Give your answer in the form \(\sqrt{a}\mbox{ m}\), where \(a\in\mathbb{N}\).
\(\sqrt{3}\mbox{ m}\)
\begin{align}\tan60^{\circ}=\frac{3}{|CA|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sqrt{3}=\frac{3}{|CA|}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|CA|&=\frac{3}{\sqrt{3}}\\&=\sqrt{3}\mbox{ m}\end{align}
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2015 Paper 2 Question 5
(a) Prove that \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\).
The answer is already in the question!
\begin{align}\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan(A-(-B))=\frac{\tan A-\tan (-B)}{1+\tan A\tan (-B)}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan(A+B)=\frac{\tan A+\tan (B)}{1-\tan A\tan B}\end{align}
where we have used \(\tan(-B)=-\tan B\).
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(b) Find all the values of \(x\) for which \(\sin(3x)=\dfrac{\sqrt{3}}{2}\), \(0\leq x\leq 360\), \(x\) in degrees.
\(x=20^{\circ},40^{\circ},140^{\circ},160^{\circ}, 260^{\circ}, 280^{\circ}\)