L.C. MATHS

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## Question 1

An experiment consists of throwing two fair, standard, six-sided dice and noting the sum of the two numbers thrown. If the sum is $$9$$ or greater it is recorded as a “win” ($$W$$). If the sum is $$8$$ or less it is recorded as a “loss” ($$L$$).

(a) Complete the table below to show all possible outcomes of the experiment.

$$t$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$

$$1$$

$$L$$

$$1$$

$$1$$

$$1$$

$$1$$

$$W$$

$$1$$

$$t$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$W$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$W$$

$$W$$

$$1$$

$$L$$

$$L$$

$$L$$

$$W$$

$$W$$

$$W$$

$$1$$

$$L$$

$$L$$

$$W$$

$$W$$

$$W$$

$$W$$

Solution
$$t$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$L$$

$$W$$

$$1$$

$$L$$

$$L$$

$$L$$

$$L$$

$$W$$

$$W$$

$$1$$

$$L$$

$$L$$

$$L$$

$$W$$

$$W$$

$$W$$

$$1$$

$$L$$

$$L$$

$$W$$

$$W$$

$$W$$

$$W$$

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(b)

(i) Find the probability of a win on one throw of the two dice.

(ii) Find the probability that each of 3 successive throws of the two dice results in a loss.

(i) $$\dfrac{5}{18}$$

(ii) $$0.3767$$

Solution

(i)

\begin{align}P(W)&=\frac{10}{36}\\&=\frac{5}{18}\end{align}

(ii)

\begin{align}P(L,L,L)&=\left(\frac{26}{36}\right)^2\\&\approx0.3767\end{align}

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(c) The experiment is repeated until a total of $$3$$ wins occur. Find the probability that the third win occurs on the tenth throw of the two dice. Give your answer correct to four decimal places.

$$0.0791$$

Solution

\begin{align}P&={9\choose2}\times\left(\frac{5}{18}\right)^2\times\left(\frac{13}{18}\right)^7\times\frac{5}{18}\\&\approx0.0791\end{align}

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## Question 2

A survey of $$100$$ shoppers, randomly selected from a large number of Saturday supermarket shoppers, showed that the mean shopping spend was €$$90.45$$. The standard deviation of this sample was €$$20.73$$.

(a) Find a $$95\%$$ confidence interval for the mean amount spent in a supermarket on that Saturday.

$$86.39\mbox{ euro}<\mu<94.51\mbox{ euro}$$

Solution

\begin{align}\bar{x}-1.96\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+1.96\frac{\sigma}{\sqrt{n}}\end{align}

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

\begin{align}90.45-1.96\frac{20.73}{\sqrt{100}}<\mu<90.45+1.96\frac{20.73}{\sqrt{100}}\end{align}

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

\begin{align}86.39\mbox{ euro}<\mu<94.51\mbox{ euro}\end{align}

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(b) A supermarket has claimed that the mean amount spent by shoppers on a Saturday is €$$94$$.
Based on the survey, test the supermarket’s claim using a $$5\%$$ level of significance. Clearly state your null hypothesis, your alternative hypothesis, and your conclusion.

The meant amount spent is $$94\mbox{ euro}$$.

Solution

Null hypothesis: Meant amount spent is $$94\mbox{ euro}$$.

Alternative hypothesis: Meant amount spent is not $$94\mbox{ euro}$$.

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma}\\&=\frac{90.45-94}{2.073}\\&\approx-1.71\end{align}

Conclusion: As $$-1.71>-1.96$$, the meant amount spent is $$94\mbox{ euro}$$.

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(c) Find the $$p$$-value of the test you performed in part (b) above and explain what this value represents in the context of the question.

$$p\mbox{-value}=0.0872$$.

If the mean amount spent was €94, then the probability that the sample mean would be €$$90.45$$ by chance is $$8.72\%$$.

Solution

\begin{align}p\mbox{-value}&=2(1-P(z<1.71)\\&=2(1-0.9564)\\&=0.0872\end{align}

If the mean amount spent was €94, then the probability that the sample mean would be €$$90.45$$ by chance is $$8.72\%$$.

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## Question 3

(a) The co-ordinates of two points are $$A(4,-1)$$ and $$B(7,t)$$.

The line $$l_1:3x-4y-12=0$$ is perpendicular to $$AB$$. Find the value of $$t$$.

$$t=-5$$

Solution

\begin{align}3x-4y-12=0\end{align}

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

\begin{align}y=\frac{3}{4}x-3\end{align}

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

\begin{align}m_{l_1}=\frac{3}{4}\end{align}

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

\begin{align}m_{AB}=-\frac{4}{3}\end{align}

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

\begin{align}\frac{t-(-1)}{7-4}=-\frac{4}{3}\end{align}

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

\begin{align}\frac{t+1}{3}=-\frac{4}{3}\end{align}

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

\begin{align}t+1=-4\end{align}

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

\begin{align}t=-5\end{align}

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(b) Find, in terms of $$k$$, the distance between the point $$P(10,k)$$ and $$l_1$$.

$$\dfrac{|18-4k|}{5}$$

Solution

\begin{align}d&=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\&=\frac{|3(10)-4k-12|}{\sqrt{3^2+4^2}}\\&=\frac{|18-4k|}{5}\end{align}

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(c) $$P(10,k)$$ is on a bisector of the angles between the lines $$l_1$$ and $$l_2:5x+12y-20=0$$.

(i) Find he possible values of $$k$$.

(ii) If $$k>0$$, find the distance from $$P$$ to $$l_1$$.

(i) $$k=\dfrac{3}{4}$$ or $$k=-48$$

(ii) $$3$$

Solution

(i)

\begin{align}\frac{|18-4k|}{5}=\frac{|5(10)+12k-20|}{\sqrt{5^2+12^2}}\end{align}

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

\begin{align}\frac{|18-4k|}{5}=\frac{|30+12k|}{13}\end{align}

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

\begin{align}13|18-4k|=5|30+12k|\end{align}

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

\begin{align}13(18-4k)=5(30+12k)\end{align}

or

\begin{align}13(18-4k)=-5(30+12k)\end{align}

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

\begin{align}-122k=-84\end{align}

or

\begin{align}8k=-384\end{align}

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

$$k=\dfrac{3}{4}$$ or $$k=-48$$

(ii)

\begin{align}d&=\frac{|18-4k|}{5}\\&=\frac{|18-4\left(\frac{3}{4}\right)|}{5}\\&=3\end{align}

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## Question 4

Two circles $$s$$ and $$c$$ touch internally at $$B$$, as shown.

(a) The equation of the circle $$s$$ is

\begin{align}(x-1)^2+(y+6)^2=360\end{align}

Write down:

• the co-ordinates of the centre of $$s$$
• the radius of $$s$$ in the form $$a\sqrt{10}$$, where $$a\in\mathbb{N}$$.

Centre: $$(1,-6)$$

Radius: $$6\sqrt{10}$$

Solution

Centre: $$(1,-6)$$

\begin{align}r&=\sqrt{360}\\&=\sqrt{36\times10}\\&=\sqrt{36}\times\sqrt{10}\\&=6\sqrt{10}\end{align}

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(b)

(i) The point $$K$$ is the centre of circle $$c$$.
The radius of $$c$$ is one-third the radius of $$s$$.
The co-ordinates of B are $$(7,12)$$.
Find the co-ordinates of $$K$$.

(ii) Find the equation of $$c$$.

(i) $$(5,6)$$

(ii) $$(x-5)^2+(y-6)^2=40$$

Solution

(i)

\begin{align}K&=\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)\\&=\left(\frac{(2)(7)+?(1)(1)}{2+1},\frac{(2)(12)+(1)(-6)}{2+1}\right)\\&=(5,6)\end{align}

(ii)

\begin{align}(x-5)^2+(y-6)^2=\left(\frac{6\sqrt{10}}{3}\right)^2\end{align}

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

\begin{align}(x-5)^2+(y-6)^2=40\end{align}

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(c) Find the equation of the common tangent at $$B$$.
Give your answer in the form $$ax+by+c=0$$, where $$a,b,c\in\mathbb{Z}$$.

$$x+3y-43=0$$

Solution

\begin{align}m_{AB}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{12-(-6)}{7-1}\\&=3\end{align}

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

\begin{align}m_{\perp}=-\frac{1}{3}\end{align}

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

\begin{align}y-y_1=m(x-x_1)\end{align}

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

\begin{align}y-12=-\frac{1}{3}(x-7)\end{align}

\begin{align}3y-36=-x+7\end{align}

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

\begin{align}x+3y-43=0\end{align}

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## Question 5

(a) Prove that $$\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$$.

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.

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

Solution

Reference Angle

\begin{align}A=60^{\circ}\end{align}

$\,$

\begin{align}3x=60^{\circ}+360^{\circ}n\end{align}

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

\begin{align}x=20^{\circ}+120^{\circ}n\end{align}

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

\begin{align}x=20^{\circ},140^{\circ},260^{\circ}\end{align}

$\,$

\begin{align}3x=120^{\circ}+360^{\circ}n\end{align}

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

\begin{align}x=40^{\circ}+120^{\circ}n\end{align}

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

\begin{align}x=40^{\circ},160^{\circ},280^{\circ}\end{align}

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## Question 6

(a) Construct the centroid of the triangle $$ABC$$ below. Show all construction lines.
(Where measurement is used, show all relevant measurements and calculations clearly.)

Solution
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(b) Prove that, if three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal line.

Solution

Diagram

Given

and

\begin{align}|AB|=BC|\end{align}

$\,$

To Prove

\begin{align}|DE|=|EF|\end{align}

$\,$

Construction

Draw $$AF’\parallel DF$$ as shown.

Draw $$F’A’\parallel CA$$ as shown.

$\,$

Proof

$$|B’F’|=|BC|$$ (parallelogram)

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

$$|B’F’|=|AB|$$ ($$|BC|=|AB|$$)

and

$$|\angle BAE’|=|\angle E’F’B’|$$ (alternate angles)

and

$$|\angle AE’B|=|\angle F’E’B’|$$ (vertically opposite angles)

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

$$\Delta ABE’$$ and $$\Delta F’B’E’$$ are congruent (ASA rule)

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

$$|AE’|=|F’E’|$$

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

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

$$|DE|=|EF|$$
since $$|AE’|=|DE|$$ and $$|F’E’|=|FE|$$

as required.

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## Question 7

A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale).
The sides of the plate, $$HK$$ and $$PQ$$, are tangential to each circle.
The larger circle has centre $$A$$ and radius $$4r\mbox{ cm}$$.
The smaller circle has centre $$B$$ and radius $$r\mbox{ cm}$$.
The length of $$[HK]$$ is $$8r\mbox{ cm}$$ and $$|AB|=20\sqrt{73}\mbox{ cm}$$.

(a) Find $$r$$, the radius of the smaller circle. (Hint: Draw $$BT\parallel KH$$, $$T\in AH$$.)

$$20\mbox{ cm}$$

Solution

\begin{align}|AT|^2+|BT|^2=|AB|^2\end{align}

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

\begin{align}(3r)^2+(8r)^2=(20\sqrt{73})^2\end{align}

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

\begin{align}9r^2+64r^2=29{,}200\end{align}

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

\begin{align}73r^2=29{,}200\end{align}

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

\begin{align}r&=\sqrt{\frac{29{,}200}{73}}\\&=20\mbox{ cm}\end{align}

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(b) Find the area of the quadrilateral $$ABKH$$.

$$8{,}000\mbox{ cm}^2$$

Solution

\begin{align}A&=|BT|\times|BK|+\frac{1}{2}|AT|\times|BT|\\&=(8r)(r)+\frac{1}{2}(3r)(8r)\\&=8r^2+12r^2\\&=20r^2\\&=20(20^2)\\&=8{,}000\mbox{ cm}^2\end{align}

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(c)

(i) Find $$|\angle HAP|$$, in degrees, correct to one decimal place.

(ii) Find the area of the machine part, correct to the nearest $$2\mbox{ cm}$$.

(i) $$138.9^{\circ}$$

(ii) $$28{,}833\mbox{ cm}^2$$

Solution

(i)

\begin{align}|\angle HAP|&=2\times|\angle HAB|\\&=2\tan^{-1}\left(\frac{8r}{3r}\right)\\&=2\tan^{-1}\left(\frac{8}{3}\right)\\&\approx138.9^{\circ}\end{align}

(ii)

\begin{align}A&=2(8000)+\pi (80^2)\left(\frac{360^{\circ}-138.9^{\circ}}{360^{\circ}}\right)+\pi (20^2)\left(\frac{138.9^{\circ}}{360^{\circ}}\right)\\&\approx28{,}833\mbox{ cm}^2\end{align}

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## Question 8

In basketball, players often have to take free throws. When Michael takes his first free throw in any game, the probability that he is successful is $$0.7$$.
For all subsequent free throws in the game, the probability that he is successful is:

• $$0.8$$ if he has been successful on the previous throw
• $$0.6$$ if he has been unsuccessful on the previous throw.

(a) Find the probability that Michael is successful ($$S$$) with all three of his first three free throws in a game.

$$0.448$$

Solution

\begin{align}P(S,S,S)&=(0.7)(0.8)(0.8)\\&=0.448\end{align}

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(b) Find the probability that Michael is unsuccessful ($$U$$) with his first two free throws and successful with the third.

$$0.072$$

Solution

\begin{align}P(U,U,S)&=(0.3)(0.4)(0.6)\\&=0.072\end{align}

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(c) List all the ways that Michael could be successful with his third free throw in a game and hence find the probability that Michael is successful with his third free throw.

$$S,S,S$$ and $$S,U,S$$ and $$U,S,S$$ and $$U,U,S$$

$$P=0.748$$

Solution

\begin{align}S,S,S&&S,U,S&&U,S,S&&U,U,S\end{align}

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

\begin{align}P&=(0.7)(0.8)(0.8)+(0.7)(0.2)(0.6)+(0.3)(0.6)(0.8)+(0.3)(0.4)(0.6)\\&=0.748\end{align}

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(d)

(i) Let $$p_n$$ be the probability that Michael is successful with his $$n$$th free throw in the game (and hence $$(1-p_n)$$ is the probability that Michael is unsuccessful with his $$n$$th free throw).
Show that $$p_{n+1}=0.6+0.2p_n$$.

(ii) Assume that $$p$$ is Michael’s success rate in the long run; that is, for large values of $$n$$, we have $$p_{n+1}\approx p_n\approx p$$.
Using the result from part (d) (i) above, or otherwise, show that $$p=0.75$$.

Solution

(i)

\begin{align}p_{n+1}&=P(S,S)+P(U,S)\\&=0.8p_n+0.6(1-p_n)\\&=0.6+0.2p_n\end{align}

as required.

(ii)

\begin{align}p\approx0.6+0.2p\end{align}

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

\begin{align}0.8p=0.6\end{align}

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

\begin{align}p&=\frac{0.6}{0.8}\\&=0.75\end{align}

as required.

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(e) For all positive integers $$n$$, let $$a_n=p-p_n$$, where $$p=0.75$$ as above.

(i) Use the ratio $$\dfrac{a_{n+1}}{a_n}$$ to show that $$a_n$$ is a geometric sequence with common ratio $$\dfrac{1}{5}$$.

(ii) Find the smallest value of $$n$$ for which $$p-p_n<0.00001$$.

(ii) $$n=7$$

Solution

(i)

\begin{align}\frac{a_{n+1}}{a_n}&=\frac{p-p_{n+1}}{p-p_n}\\&=\frac{0.75-(0.6+0.2p_n)}{0.75-p_n}\\&=\frac{0.15-0.2p_n}{0.75-p_n}\\&=\frac{0.15-0.2p_n}{5(0.15-0.2p_n)}\\&=\frac{1}{5}\end{align}

as required.

(ii)

\begin{align}a_1&=p-p_1\\&=0.75-0.7=0.05\end{align}

\begin{align}a=0.05&&r=\frac{1}{5}\end{align}

\begin{align}a_n<0.00001\end{align}

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

\begin{align}ar^{n-1}<0.00001\end{align}

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

\begin{align}(0.05)\left(\frac{1}{5}\right)^{n-1}<0.00001\end{align}

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

\begin{align}\left(\frac{1}{5}\right)^{n-1}<0.0002\end{align}

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

\begin{align}\ln\left(\frac{1}{5}\right)^{n-1}<\ln0.0002\end{align}

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

\begin{align}(n-1)\ln\left(\frac{1}{5}\right)<\ln0.0002\end{align}

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

\begin{align}n>1+\frac{\ln0.0002}{\ln\left(\frac{1}{5}\right)}\end{align}

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

\begin{align}n>6.29…\end{align}

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

\begin{align}n=7\end{align}

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(f) You arrive at a game in which Michael is playing. You know that he has already taken many free throws, but you do not know what pattern of success he has had.

(i) Based on this knowledge, what is your estimate of the probability that Michael will be successful with his next free throw in the game?

(ii) Why would it not be appropriate to consider Michael’s subsequent free throws in the game as a sequence of Bernoulli trials?

(i) $$0.75$$

(ii) Each subsequent free throw is not an independent event.

Solution

(i) $$0.75$$

(ii) Each subsequent free throw is not an independent event.

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## Question 9

(a) Joan is playing golf. She is $$150\mbox{ m}$$ from the centre of a circular green of diameter $$30\mbox{ m}$$. The diagram shows the range of directions in which Joan can hit the ball so that it could land on the green. Find $$\alpha$$, the measure of the angle of this range of directions. Give your answer, in degrees, correct to one decimal place.

$$11.5^{\circ}$$

Solution

\begin{align}\sin\left(\frac{\alpha}{2}\right)=\frac{15}{150}\end{align}

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

\begin{align}\frac{\alpha}{2}=5.739…^{\circ}\end{align}

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

\begin{align}\alpha&=2(5.739…^{\circ})\\&\approx11.5^{\circ}\end{align}

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(b) At the next hole, Joan, at $$T$$, attempts to hit the ball in the direction of the hole $$H$$. Her shot is off target and the ball lands at $$A$$, a distance of $$190$$ metres from $$T$$, where $$|\angle ATH=18^{\circ}$$.
$$|TH|$$ is $$385$$ metres. Find $$|AH|$$, the distance from the ball to the hole, correct to the nearest metre.

$$213\mbox{ m}$$

Solution

\begin{align}|AH|^2=|AT|^2+|TH|^2-2|AT||TH|\cos|\angle ATH|\end{align}

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

\begin{align}|AH|^2&=(190)^2+(385)^2-2(190)(385)\cos18^{\circ}\\&=45{,}185.4317…\end{align}

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

\begin{align}|AH|&=\sqrt{45{,}185.4317…}\\&\approx213\mbox{ m}\end{align}

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(c) At another hole, where the ground is not level, Joan hits the ball from $$K$$, as shown. The ball lands at $$B$$. The height of the ball, in metres, above the horizontal line $$OB$$ is given by

\begin{align}h=-6t^2+22t+8\end{align}

where $$t$$ is the time in seconds after the ball is struck and h is the height of the ball.

(i) Find the height of $$K$$ above $$OB$$.

(ii) The horizontal speed of the ball over the straight distance $$[OB]$$ is a constant $$38\mbox{ m s}^{-1}$$.
Find the angle of elevation of $$K$$ from $$B$$, correct to the nearest degree.

(i) $$8\mbox{ m}$$

(ii) $$3^{\circ}$$

Solution

(i)

\begin{align}h(0)&=-6(0^2)+22(0)+8\\&=8\mbox{ m}\end{align}

(ii)

\begin{align}h(t)=0\end{align}

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

\begin{align}-6t^2+22t+8=0\end{align}

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

\begin{align}3t^2-11t-4=0\end{align}

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

\begin{align}(3t+1)(t-4)=0\end{align}

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

\begin{align}t=4\mbox{ s}\end{align}

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

\begin{align}|OB|&=(38)(4)\\&=152\mbox{ m}\end{align}

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

\begin{align}|\angle KBO|&=\tan^{-1}\left(\frac{8}{152}\right)\\&\approx3^{\circ}\end{align}

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(d) At a later hole, Joan’s first shot lands at the point $$G$$, on ground that is sloping downwards, as shown. A vertical tree, $$[CE]$$, $$25$$ metres high, stands between $$G$$ and the hole. The distance, $$|GC|$$, from the ball to the bottom of the tree is also $$25$$ metres.
The angle of elevation at $$G$$ to the top of the tree, $$E$$, is $$\theta$$, where $$\theta=\tan^{-1}\dfrac{1}{2}$$.
The height of the top of the tree above the horizontal, $$GD$$, is $$h$$ metres and $$|GD|=d$$ metres.

(i) Write $$d$$ and $$|CD|$$ in terms of $$h$$.

(ii) Hence, or otherwise, find $$h$$.

(i) $$d=2h$$ and $$|CD|=25-h$$

(ii) $$h=10\mbox{ m}$$

Solution

(i)

\begin{align}\tan\theta=\frac{1}{2}\end{align}

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

\begin{align}\frac{h}{d}=\frac{1}{2}\end{align}

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

\begin{align}d=2h\end{align}

and

\begin{align}|CD|=25-h\end{align}

(ii)

\begin{align}|CD|^2+d^2=25^2\end{align}

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

\begin{align}(25-h)^2+(2h)^2=625\end{align}

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

\begin{align}625-50h+h^2+4h^2=625\end{align}

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

\begin{align}5h^2-50h=0\end{align}

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

\begin{align}h(5h-50)=0\end{align}

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

\begin{align}h=10\mbox{ m}\end{align}

Video Walkthrough