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## Question 1

When Conor rings Ciara’s house, the probability that Ciara answers the phone is $$\dfrac{1}{5}$$.

(a) Conor rings Ciara’s house once every day for $$7$$ consecutive days. Find the probability that she will answer the phone on the $$2$$nd, $$4$$th, and $$6$$th days but not on the other days.

$$\dfrac{256}{78{,}125}$$

Solution

\begin{align}P&=\frac{4}{5}\times\frac{1}{5}\times\frac{4}{5}\times\frac{1}{5}\times\frac{4}{5}\times\frac{1}{5}\times\frac{4}{5}\\&=\frac{256}{78{,}125}\end{align}

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(b) Find the probability that she will answer the phone for the $$4$$th time on the $$7$$th day.

$$\dfrac{1280}{78{,}125}$$

Solution

\begin{align}P&={6\choose3}\times\left(\frac{1}{5}\right)^3\times\left(\frac{4}{5}\right)^3\times\frac{1}{5}\\&=\frac{1280}{78{,}125}\end{align}

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(c) Conor rings her house once every day for $$n$$ days. Write, in terms of $$n$$, the probability that Ciara will answer the phone at least once.

$$1-\left(\dfrac{4}{5}\right)^n$$

Solution

\begin{align}P(\mbox{at least once})&=1-P(\mbox{never})\\&=1-\left(\frac{4}{5}\right)^n\end{align}

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(d) Find the minimum value of $$n$$ for which the probability that Ciara will answer the phone at least once is greater than $$99\%$$.

$$n=21$$

Solution

\begin{align}1-\left(\frac{4}{5}\right)^n>0.99\end{align}

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

\begin{align}\left(\frac{4}{5}\right)^n<0.01\end{align}

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

\begin{align}n\ln\left(\frac{4}{5}\right)<\ln(0.01)\end{align}

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

\begin{align}n<\frac{\ln(0.01)}{\ln\left(\frac{4}{5}\right)}\end{align}

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

\begin{align}n<20.637…\end{align}

Therefore, the minimum value of $$n$$ is $$21$$.

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## Question 2

An experiment measures the fuel consumption at various speeds for a particular model of car.
The data collected are shown in Table $$1$$ below.

Speed (km/hour) $$40$$ $$48$$ $$56$$ $$64$$ $$88$$ $$96$$ $$112$$

Fuel consumption (km/litre)

$$21$$

$$16$$

$$18$$

$$16$$

$$13$$

$$11$$

$$9$$

(a) Find the correlation coefficient of the data in Table $$1$$, correct to $$3$$ decimal places.

$$r=-0.957$$

Solution

$$r=-0.957$$

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(b) Plot the points from the table on the grid below and draw the line of best fit (by eye).

Solution
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(c) The slope of the line of best fit is found to be $$-0.15$$.
What does this value represent in the context of the data?

$$0.15$$ is the rate at which fuel consumption is decreasing as the speed increases.

Solution

$$0.15$$ is the rate at which fuel consumption is decreasing as the speed increases.

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(d) Mary drove from Cork to Dublin at an average speed of $$96\mbox{ km/h}$$.
Jane drove the same journey at an average speed of $$112\mbox{ km/h}$$.
Each travelled $$260\mbox{ km}$$ and paid $$132.9$$ cents per litre for the fuel.
Both used the model of car used to generate the data in Table $$1$$.

(i) Find how much longer it took Mary to complete the journey.

(ii) Based on the data in Table $$1$$ and their average speeds, find how much more Jane spent on fuel during the course of this journey.

(i) $$23\mbox{ min}$$

(ii) $$6.98\mbox{ euro}$$

Solution

(i)

\begin{align}t_M-t_J&=\frac{d_M}{s_M}-\frac{d_J}{s_J}\\&=\frac{260}{96}-\frac{260}{112}\\&=0.3689..\mbox{ hr}\\&\approx23\mbox{ min}\end{align}

(ii)

\begin{align}\left(\frac{260}{9}\times1.329\right)-\left(\frac{260}{11}\times1.329\right)\approx6.98\mbox{ euro}\end{align}

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## Question 3

$$ABC$$ is a triangle where the co-ordinates of $$A$$ and $$C$$ are $$(0,6)$$ and $$(4,2)$$ respectively.
$$G\left(\dfrac{2}{3},\dfrac{4}{3}\right)$$ is the centroid of the triangle $$ABC$$.
$$AG$$ intersects $$BC$$ at the point $$P$$.
$$|AG|:|GP|=2:1$$.

(a) Find the co-ordinates of $$P$$.

$$(1,-1)$$

Solution

\begin{align}\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)=\left(\frac{2}{3},\frac{4}{3}\right)\end{align}

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

\begin{align}\left(\frac{(1)(0)+(2)x_2}{1+2},\frac{(1)(6)+(2)y_2}{1+2}\right)=\left(\frac{2}{3},\frac{4}{3}\right)\end{align}

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

\begin{align}\left(\frac{2x_2}{3},\frac{6+2y_2}{3}\right)=\left(\frac{2}{3},\frac{4}{3}\right)\end{align}

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

\begin{align}2x_2=2\end{align}

\begin{align}6+2y_2=4\end{align}

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

\begin{align}x_2=1\end{align}

\begin{align}y_2=-1\end{align}

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

\begin{align}P:(1,-1)\end{align}

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(b) Find the co-ordinates of $$B$$.

$$B:(-2,-4)$$

Solution

\begin{align}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=(1,-1)\end{align}

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

\begin{align}\left(\frac{4+x_2}{2},\frac{2+y_2}{2}\right)=(1,-1)\end{align}

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

\begin{align}\frac{4+x_2}{2}=1\end{align}

\begin{align}\frac{2+y_2}{2}=-1\end{align}

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

\begin{align}x_2&=(1)(2)-4\\&=-2\end{align}

\begin{align}y_2&=(-1)(2)-2\\&=-4\end{align}

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

\begin{align}B:(-2,-4)\end{align}

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(c) Prove that $$C$$ is the orthocentre of the triangle $$ABC$$.

Solution

\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{6-2}{0-4}\\&=-1\end{align}

and

\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-4-2}{-2-4}\\&=1\end{align}

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

\begin{align}m_{AC}\times m_{BC}&=(-1)(1)\\&=-1\end{align}

Therefore, as the two lines are perpendicular to each other, $$C$$ must be the orthocentre of the triangle.

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## Question 4

$$A(0,0)$$, $$B(6.5,0)$$ and $$C(10,7)$$ are three points on a circle.

(a) Find the equation of the circle.

$$x^2+y^2-6.5x-12y=0$$

Solution

\begin{align}x^2+y^2+2gx+2fy+c=0\end{align}

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

\begin{align}0^2+0^2+2g(0)+2f(0)+c=0\end{align}

and

\begin{align}6.5^2+0^2+2g(6.5)+2f(0)+c=0\end{align}

and

\begin{align}10^2+7^2+2g(10)+2f(7)+c=0\end{align}

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

\begin{align}c=0\end{align}

\begin{align}42.25+13g+c=0\end{align}

\begin{align}149+20g+14f+c=0\end{align}

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

\begin{align}42.25+13g=0\end{align}

\begin{align}149+20g+14f=0\end{align}

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

\begin{align}g=-3.25\end{align}

\begin{align}149+20g+14f=0\end{align}

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

\begin{align}149+20(-3.25)+14f=0\end{align}

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

\begin{align}f&=\frac{20(3.25)-149}{14}\\&=-6\end{align}

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

\begin{align}x^2+y^2-6.5x-12y=0\end{align}

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(b) Find $$|\angle BCA|$$. Give your answer in degrees, correct to $$2$$ decimal places.

$$28.44^{\circ}$$

Solution

\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-7}{6.5-10}\\&=2\end{align}

and

\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-7}{0-10}\\&=\frac{7}{10}\end{align}

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

\begin{align}\tan|\angle BCA|&=\pm\frac{m_{BC}-m_{AC}}{1+m_{BC}m_{AC}}\\&=\pm\frac{2-\frac{7}{10}}{1+(2)(\left(\frac{7}{10}\right)}\\&=\frac{\frac{13}{10}}{\frac{24}{10}}\\&=\frac{13}{24}\end{align}

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

\begin{align}|\angle BCA|&=\tan^{-1}\left(\frac{13}{24}\right)\\&\approx28.44^{\circ}\end{align}

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## Question 5

$$ABCD$$ is a rectangle.

$$F\in[AB]$$, $$G\in[BC]$$, $$[FD]\cap[AG]={E}$$, and $$FD \perp AG$$.

$$|AE|=12\mbox{ cm}$$, $$|EG|=27\mbox{ cm}$$, and $$|FE|=5\mbox{ cm}$$.

(a) Prove that $$\Delta AFE$$ and $$\Delta DAE$$ are similar (equiangular).

Solution

$$|\angle AED|$$ is a right angle.

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

$$|\angle ADE|+|\angle EAD|=90^{\circ}$$ (remaining angles)

and

$$|\angle FAE|+|\angle EAD|=90^{\circ}$$ (rectangle corner)

Comparing the to equations, we can conclude that $$|\angle ADE|=|\angle FAE|$$. As both triangle also contain a right angle, their third angles are also the same.

Therefore, both triangles are similar.

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(b) Find $$|AD|$$.

$$31.2\mbox{ cm}$$

Solution

\begin{align}|AF|&=\sqrt{5^2+12^2}\\&=13\end{align}

and

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

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

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(c) $$\Delta AFE$$ and $$\Delta AGB$$ are similar. Show that $$|AB|=36\mbox{ cm}$$.

$$36\mbox{ cm}$$

Solution

\begin{align}\frac{|AG|}{|AF|}=\frac{|AB|}{|AE|}\end{align}

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

\begin{align}\frac{39}{13}=\frac{|AB|}{12}\end{align}

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

\begin{align}|AB|&=12\times\frac{39}{13}\\&=36\mbox{ cm}\end{align}

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(d) Find the area of the quadrilateral $$GCDE$$.

$$680.4\mbox{ cm}^2$$

Solution

\begin{align}A_{GCDE}&=A_{ABCD}-A_{ABG}-A_{AFD}+A_{AFE}\\&=(31.2)(36)-\frac{1}{2}(36)(15)-\frac{1}{2}(31.2)(13)+\frac{1}{2}(5)(12)\\&=680.4\mbox{ cm}^2\end{align}

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## 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}

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## Question 7

Two solid cones, each of radius $$R\mbox{ cm}$$ and height $$R\mbox{ cm}$$ are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below.

(a) Show that the capacity (volume) of the empty space in the cylinder is equal to the capacity of an empty sphere of radius $$R\mbox{ cm}$$ (Figure 2).

Solution

\begin{align}V&=\pi r^2h_1-2\left(\frac{1}{3}\pi r^2h_2\right)\\&=\pi R^2(2R)-2\left(\frac{1}{3}\pi R^2(R)\right)\\&=2\pi R^3-\frac{2}{3}\pi R^3\\&=\frac{4}{3}\pi R^3\end{align}

as required.

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(b) In the remainder of this question, $$R=12\mbox{ cm}$$. Water is poured into both the cylinder and the
sphere to a depth of 6 cm as shown below (Figure 3 and Figure 4 respectively).

(i) Find $$|AB|$$, the radius of the circular surface of the water in the sphere (Figure 4).
Give your answer in the form ܽ$$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

(ii) Find $$|CD|$$, the radius of the cone at water level, as shown in Figure 3.

(iii) Verify that the area of the surface of the water in the sphere is equal to the area of the surface of the water in the cylinder.

(i) $$6\sqrt{3}\mbox{ cm}$$

(ii) $$6\mbox{ cm}$$

Solution

(i)

\begin{align}|AB|&=\sqrt{12^2-(12-6)^2}\\&=\sqrt{108}\\&=6\sqrt{3}\mbox{ cm}\end{align}

(ii)

\begin{align}\frac{r}{12}=\frac{6}{12}\end{align}

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

\begin{align}r=6\mbox{ cm}\end{align}

(iii)

\begin{align}A_C&=\pi(12^2)-\pi(6^2)\\&=108\pi\end{align}

and

\begin{align}A_S&=\pi(6\sqrt{3})^2\\&=108\pi\end{align}

as required.

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(c) The mathematician Cavalieri discovered that, at the same depth, the volume of water in the available space in the cylinder is equal to the volume of water in the sphere.
Use this discovery to find the volume of water in the sphere when the depth is $$6\mbox{ cm}$$.
Give your answer in terms of $$\pi$$.

$$360\pi\mbox{ cm}^3$$

Solution

\begin{align}V&=\pi(12^2)(6)-\left[\frac{1}{3}\pi(12^2)(12)-\frac{1}{3}\pi(6^2)(6)\right]\\&=360\pi\mbox{ cm}^3\end{align}

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## Question 8

(a) In $$2015$$, in a particular country, the weights of $$15$$ year olds were normally distributed with a mean of $$63.5\mbox{ kg}$$ and a standard deviation of $$10\mbox{ kg}$$.

(i) In $$2015$$, Mariska was a $$15$$ year old in that country. Her weight was $$50\mbox{ kg}$$.
Find the percentage of $$15$$ year olds in that country who weighed more than Mariska.

(ii) In $$2015$$, Kamal was a $$15$$ year old in that country.
$$1.5\%$$ of $$15$$ year olds in that country were heavier than Kamal.
Find Kamal’s weight.

(iii) In $$2016$$, $$150$$ of the $$15$$ year olds in that country were randomly selected and their weights recorded. It was found that their weights were normally distributed with a mean weight of $$62\mbox{ kg}$$ and a standard deviation of $$10\mbox{ kg}$$. Test the hypothesis, at the $$5\%$$ level of significance, that the mean weight of $$15$$ year olds, in that country, had not changed from $$2015$$ to $$2016$$. State the null hypothesis and your alternative hypothesis.
Give your conclusion in the context of the question.

(i) $$0.9115$$

(ii) $$85.2\mbox{ kg}$$

(iii) The mean weight has not changed.

Solution

(i)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{50-63.5}{10}\\&=-1.35\end{align}

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

\begin{align}P(z<1.35)=0.9115\end{align}

(ii)

\begin{align}P(x<z)&=1-0.015\\&=0.985\end{align}

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

\begin{align}z=2.17\end{align}

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

\begin{align}\frac{x-\mu}{\sigma}=2.17\end{align}

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

\begin{align}\frac{x-63.5}{10}=2.17\end{align}

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

\begin{align}x&=(10)(2.17)+63.5\\&=85.2\mbox{ kg}\end{align}

(iii)

Null hypothesis: Mean weight had not changed.

Alternative hypothesis: Mean weight had changed.

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{62-63.5}{10/\sqrt{150}}\\&=-1.837…\end{align}

Conclusion:

As $$-1.837…>-1.96$$, the mean weight has not changed.

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(b) In Galway, rain falls in the morning on $$\dfrac{1}{3}$$ of the school days in the year.

When it is raining the probability of heavy traffic is $$\dfrac{1}{2}$$.

When it is not raining the probability of heavy traffic is $$\dfrac{1}{4}$$.

When it is raining and there is heavy traffic, the probability of being late for school is $$\dfrac{1}{2}$$.

When it is not raining and there is no heavy traffic, the probability of being late for school is $$\dfrac{1}{8}$$.

In any other situation the probability of being late for school is $$\dfrac{1}{5}$$.

Some of this information is shown in the tree diagram below.

(i) Write the probability associated with each branch of the tree diagram and the probability of each outcome into the blank boxes provided.
Give each answer in the form $$\dfrac{a}{b}$$, where $$a,b\in\mathbb{N}$$.

(ii) On a random school day in Galway, find the probability of being late for school.

(iii) On a random school day in Galway, find the probability that it rained in the morning, given that you were late for school.

(i)

(ii) $$\dfrac{17}{80}$$

(iii) $$\frac{28}{51}$$

Solution

(i)

(ii)

\begin{align}P&=\frac{1}{12}+\frac{1}{30}+\frac{2}{60}+\frac{6}{96}\\&=\frac{17}{80}\end{align}

(iii)

\begin{align}P(R|L)&=\frac{P(R\cap L)}{P(L)}\\&=\frac{\frac{1}{12}+\frac{1}{30}}{\frac{17}{80}}\\&=\frac{28}{51}\end{align}

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## Question 9

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.

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}

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(f) If the tree was equally likely to fall in any direction, find the probability that it would hit the bank at Conor’s side, when it falls.
Give your answer as a percentage, correct to $$1$$ decimal place

$$30.4\%$$