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...

Area of a Triangle

Sine/Cosine Rules

3D Problems

Trigonometric Equations

Trigonometric Identities

Has not appeared

Compound Angle Formulae

Double-Angle Formulae

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.

$$3.85\mbox{ minutes}$$

Solution

\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}

Cosine is negative in 2nd quadrant (1st time) and 3rd quadrant (2nd time).

$\,$

\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}

Video Walkthrough

## 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) $$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

Solution

(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}

Video Walkthrough
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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 $$|AC|$$.

$$10$$

Solution

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}

Video Walkthrough

## 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 $$A_2=134\mbox{ m}^2$$

Solution

\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}\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

$$93\mbox{ m}$$

Solution

\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}

Video Walkthrough
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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) $$2.6^{\circ}$$

Solution

(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) $$4.0\mbox{ km}$$

Solution

(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}

Video Walkthrough

## 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) $$\cos\theta=\dfrac{3}{5}$$

Solution

(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}

Video Walkthrough
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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 $$B=330^{\circ}$$

Solution

\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.

$\,$

\begin{align}B&=180^{\circ}-30^{\circ}\\&=150^{\circ}\end{align}

$\,$

\begin{align}B&=360^{\circ}-30^{\circ}\\&=330^{\circ}\end{align}

Video Walkthrough

## 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!

Solution

\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.

Video Walkthrough
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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.

$$50.1\mbox{ km}^2$$

Solution

\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.

$$26\mbox{ m}$$

Solution

\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}

Video Walkthrough

## 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!

Solution

(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.

Video Walkthrough
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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) $$58{,}646\mbox{ Volts/sec}$$

Solution

(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}

Video Walkthrough

## 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) $$50\mbox{ square units}$$

Solution

(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}

Video Walkthrough

## 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 $$\dfrac{11\pi}{3}$$

Solution

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.

$\,$

\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}

$\,$

\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.

$$4.4\mbox{ cm}$$

Solution

\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}\downarrow\end{align}

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

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

\begin{align}|BC|&=|BO|-|CO|\\&=7-2.57…\\&\approx4.4\mbox{ cm}\end{align}

Video Walkthrough

## 2019 Paper 2 Question 4

(a) Show that $$\cos2\theta=1-2\sin^2\theta$$.

The answer is already in the question!

Solution

\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.

Video Walkthrough
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(b) Find the cosine of the acute angle between two diagonals of a cube.

$$\dfrac{1}{3}$$

Solution

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.

$$54.4\mbox{ m}$$

Solution

\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.

$$30.75^{\circ}$$

Solution

\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

$$806.65\mbox{ m}^2$$

Solution

\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) $$11.3\mbox{ m}$$

Solution

(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.

$$10.7\mbox{ m}$$

Solution

\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 $$x=285^{\circ}$$

Solution

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}

$\,$

\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 $$x=255^{\circ}$$

$\,$

\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 $$y$$.

$$\dfrac{y\sqrt{4-y^2}}{2}$$

Solution

\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 $$\mbox{km}$$.

$$52\mbox{ km}$$

Solution

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 $$\mbox{km}$$.

$$24{,}091\mbox{ km}$$

Solution

\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}

Video Walkthrough

## 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 $$a\in\mathbb{N}$$.

$$|TE|=\sqrt{3}|CT|$$

Solution

\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!

Solution

\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.

$$|CT|=5.3\mbox{ m}$$

Solution

\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.

$$9.2\mbox{ m}$$

Solution

\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.

$$54.7^{\circ}$$

Solution

\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}

Video Walkthrough

## 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!

Solution

(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!

Solution

\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 $$a,b\in\mathbb{N}$$.

$$\pm\dfrac{\sqrt{5}}{3}$$

Solution

\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) $$15\mbox{ m}^2$$

Solution

(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 $$a\in\mathbb{N}$$.

$$\sqrt{3}\mbox{ m}$$

Solution

\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!

Solution

\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.

$$x=20^{\circ},40^{\circ},140^{\circ},160^{\circ}, 260^{\circ}, 280^{\circ}$$