Course Content
Higher Level (by year)
0/17
Higher Level (by topic)
0/13
Ordinary Level (by year)
0/17
Ordinary Level (by topic)
0/12
Past Papers

## Question 1

(a) A class consists of $$12$$ boys and $$8$$ girls.

(i) Two students are selected at random from the class. What is the probability that the two students selected will be a boy and a girl in any order?

(ii) Four students are selected, one at a time, at random from the class.
What is the probability that the first three students selected will be boys and the fourth will be a girl?

(i) $$\dfrac{48}{95}$$

(ii) $$\dfrac{88}{969}$$

Solution

(i)

\begin{align}P&=\frac{12}{20}\times\frac{8}{19}+\frac{8}{20}\times\frac{12}{19}\\&=\frac{48}{95}\end{align}

(ii)

\begin{align}P&=\frac{12}{20}\times\frac{11}{19}\times\frac{10}{18}\times\frac{8}{17}\\&=\frac{88}{969}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) An examination paper is made up of two sections, Section $$A$$ consisting of $$7$$ questions
and Section $$B$$ consisting of $$8$$ questions. The paper contains the following instruction:

“From section $$A$$ you must answer question $$1$$ and any three other questions.
From Section $$B$$ you must also answer any four questions.”

Find how many different combinations of questions may be answered if a candidate follows
this instruction.

$$1{,}400$$

Solution

\begin{align}N&={1\choose1}\times{6\choose3}\times{8\choose4}\\&=1{,}400\end{align}

Video Walkthrough

## Question 2

(a) The line $$p$$ makes an intercept on the $$x$$-axis at $$(a,0)$$ and on the $$y$$-axis at $$(0,b)$$, where $$a,b\neq0$$.

Show that the equation of $$p$$ can be written as $$\dfrac{x}{a}+\dfrac{y}{b}=1$$.

Solution

\begin{align}m&=\frac{b-0}{0-a}\\&=-\frac{b}{a}\end{align}

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

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

\begin{align}y-0=-\frac{b}{a}(x-a)\end{align}

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

\begin{align}ay=-bx+ab\end{align}

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

\begin{align}bx+ay=ab\end{align}

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

\begin{align}\frac{x}{a}+\frac{y}{b}=1\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) The line $$l$$ has a slope $$m$$, and contains the point $$A(6,0)$$.

(i) Write the equation of the line $$l$$ in term of $$m$$.

(ii) The line $$l$$ cuts the line $$k:4x+3y=25$$ at $$P$$.
Find the co-ordinates of $$P$$ in terms of $$m$$.
Give each co-ordinate as a fraction in its simplest form.

(i) $$y=mx-6m$$

(ii) $$P:\left(\dfrac{18m+25}{3m+4},\dfrac{m}{3m+4}\right)$$

Solution

(i)

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

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

\begin{align}y-0=m(x-6)\end{align}

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

\begin{align}y=mx-6m\end{align}

(ii)

\begin{align}y=mx-6m\end{align}

\begin{align}4x+3y=25\end{align}

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

\begin{align}mx-y=6m\end{align}

\begin{align}4x+3y=25\end{align}

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

\begin{align}3mx-3y=18m\end{align}

\begin{align}4x+3y=25\end{align}

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

\begin{align}(3m+4)x=18m+25\end{align}

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

\begin{align}x=\frac{18m+25}{3m+4}\end{align}

and

\begin{align}y&=mx-6m\\&=m\left(\frac{18m+25}{3m+4}\right)-6m\\&=\frac{18m^2+25m-18m^2-24m}{3m+4}\\&=\frac{m}{3m+4}\end{align}

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

\begin{align}P:\left(\frac{18m+25}{3m+4},\frac{m}{3m+4}\right)\end{align}

Video Walkthrough

## Question 3

(a) The point $$(-2,k)$$ is on the circle $$(x-2)^2+(y-3)^2=65$$.
Find the two possible values of $$k$$, where $$k\in\mathbb{Z}$$.

$$k=-4$$ or $$k=10$$

Solution

\begin{align}(x-2)^2+(y-3)^2=65\end{align}

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

\begin{align}(-2-2)^2+(k-3)^2=65\end{align}

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

\begin{align}16+k^2-6k+9=65\end{align}

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

\begin{align}k^2-6k-40=0\end{align}

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

\begin{align}(k+4)(k-10)=0\end{align}

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

$$k=-4$$ or $$k=10$$

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) The circle $$s$$ is in the first quadrant. It touches both the $$x$$-axis and the $$y$$-axis.
The line $$t:3x-4y+6=0$$ is a tangent to $$s$$ as shown. Find the equation of $$s$$.

$$(x-1)^2+(y-1)^2=1$$

Solution

The centre of the circle is $$(r,r)$$ and the radius is $$r$$.

\begin{align}\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=r\end{align}

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

\begin{align}\frac{|3r-4r+6||}{\sqrt{3^2+4^2}}=r\end{align}

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

\begin{align}\frac{|-r+6|}{5}=r\end{align}

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

\begin{align}|-r+6|=5r\end{align}

$$-r+6=5r$$ or $$-r+6=-5r$$

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

$$r=1$$ or $$r=-\dfrac{3}{2}$$

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

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

(as we are in the first quadrant).

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

Video Walkthrough

## Question 4

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

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
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(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}

Video Walkthrough

## Question 5

(a) Construct and label the orthocentre of the triangle $$ABC$$ in the diagram below.
Show any construction lines or arcs clearly.

Solution
Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) In the diagram below $$O$$ is the centre of circle s. $$[AB]$$ is a diameter of $$s$$.
$$BE$$ is a tangent to $$s$$ at point $$B$$.
$$[CD]$$ is a chord of circle $$s$$.
$$|CD|=\dfrac{1}{2}|AB|$$ and $$CD$$ is parallel to $$AB$$.
Find, with justification, $$|\angle BEA|$$.

$$30^{\circ}$$

Solution

\begin{align}|OD|=|DC|=|OC|=r\end{align}

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

Triangle $$DOC$$ is equilateral

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

\begin{align}|\angle DOC|=60^{\circ}\end{align}

By symmetry, $$|\angle AOD|=|\angle BOC|$$ and therefore

\begin{align}|\angle AOD|&=\frac{180^{\circ}-60^{\circ}}{2}\\&=60^{\circ}\end{align}

Therefore, as triangle $$AOD$$ is isosceles:

Finally, for triangle $$ABE$$, since $$BE$$ is a tangent:

\begin{align}|\angle BEA|&=180^{\circ}-|\angle BAE|-|\angle ABE|\\&=180^{\circ}-60^{\circ}-90^{\circ}\\&=30^{\circ}\end{align}

Video Walkthrough

## Question 6

(a) Two independent events $$F$$ and $$S$$ are represented in the Venn diagram shown below.
$$P(F\backslash S)=\dfrac{1}{4}$$, $$P(F\cap S)=\dfrac{1}{5}$$, $$P(S\backslash F)=x$$, and $$P(F\cup S)’=y$$, where $$x,y\neq0$$.
Find the value of $$x$$ and the value of $$y$$.

$$x=\dfrac{11}{45}$$ and $$y=\dfrac{11}{36}$$

Solution

\begin{align}P(F\cap S)=P(F)\times P(S)\end{align}

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

\begin{align}\frac{1}{5}=\left(\frac{1}{4}+\frac{1}{5}\right)\times P(S)\end{align}

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

\begin{align}\frac{1}{5}=\frac{9}{20}\times P(S)\end{align}

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

\begin{align}P(S)&=\frac{20}{45}\\&=\frac{4}{9}\end{align}

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

\begin{align}\frac{1}{5}+x=\frac{4}{9}\end{align}

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

\begin{align}x&=\frac{4}{9}-\frac{1}{5}\\&=\frac{20-9}{45}\\&=\frac{11}{45}\end{align}

and

\begin{align}\frac{1}{4}+\frac{1}{5}+x+y=1\end{align}

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

\begin{align}\frac{1}{4}+\frac{1}{5}+\frac{11}{45}+y=1\end{align}

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

\begin{align}y&=1-\frac{1}{4}-\frac{1}{5}-\frac{11}{45}\\&=\frac{11}{36}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) In a club there are German, Irish and Spanish children only.
There are $$10$$ Spanish children.
There are twice as many Irish children as German children.

They are all in a group waiting to get on a swing.
One child will be selected at random to go first and will not re-join the group.
Then a second child will be selected at random to go next.

The probability that the first child selected will be German and that the second child selected will not be German is $$\dfrac{1}{6}$$.

Find how many children are in the club.

$$25$$

Solution

Let $$x$$ be the number of German children.

• There are $$10$$ Spanish children
• There are $$x$$ German children
• There are $$2x$$ Irish children

In total, there are therefore $$10+x+2x=10+3x$$ children.

\begin{align}\frac{x}{10+3x}\times\frac{2x+10}{9+3x}=\frac{1}{6}\end{align}

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

\begin{align}\frac{2x^2+10x}{90+30x+27x+9x^2}=\frac{1}{6}\end{align}

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

\begin{align}12x^2+60x=90+57x+9x^2\end{align}

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

\begin{align}3x^2+3x-90=0\end{align}

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

\begin{align}x^2+x-30=0\end{align}

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

\begin{align}(x+6)(x-5)=0\end{align}

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

\begin{align}x=5\end{align}

(as $$x>0$$). Therefore, the total number of children is

\begin{align}10+3x&=10+3(5)\\&=25\end{align}

Video Walkthrough

## Question 7

(a) A cattle feeding trough of uniform cross section and $$2.5\mbox{ m}$$ in length, is shown in Figure 1.
The front of the trough (segment $$ABC$$) is shown in Figure 2.
The front of the trough is a segment of a circle of radius $$0\mbox{ cm}$$.
The height of the trough, $$|DB|$$, is $$30\mbox{ cm}$$.
$$|OA|=|OC|=|OB|=90\mbox{ cm}$$. $$[OB]\perp[AC]$$.

(i) Find $$|AD|$$. Give your answer in the form $$a\sqrt{B}\mbox{ cm}$$, where $$a,b\in\mathbb{Z}$$.

(ii) Find $$|\angle DOA|$$. Give your answer in radians, correct to $$2$$ decimal places.

(iii) Find the area of the segment $$ABC$$. Give your answer in $$\mbox{m}^2$$ correct to $$2$$ decimal places.

(iv) Find the volume of the trough. Give your answer in $$\mbox{m}^3$$, correct to $$2$$ decimal places.

(i) $$30\sqrt{5}\mbox{ cm}$$

(ii) $$0.84\mbox{ rad}$$

(iii) $$0.28\mbox{ m}^2$$

(iv) $$0.70\mbox{ m}^3$$

Solution

(i)

(ii)

(iii)

\begin{align}A&=\frac{1}{2}r^2\theta-\frac{1}{2}bh\\&=\frac{1}{2}(90^2)(2\times0.84)-\frac{1}{2}(2\times30\sqrt{5})(90-30)\\&\approx0.28\mbox{ m}^2\end{align}

(iv)

\begin{align}V&=Al\\&=(0.28)(2.5)\\&=0.70\mbox{ m}^3\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) A sand timer for games is shown in the diagram.
Each half of the timer consists of a hemisphere, a cylinder of height $$3.5\mbox{ cm}$$ and a cone of height $$1.5\mbox{ cm}$$.
All of the parts have a radius of $$1.25\mbox{ cm}$$.

(i) The upper half of the timer is full of sand.
Find the volume of sand in the upper half of the timer.
Give your answer in $$\mbox{cm}^3$$ correct to $$2$$ decimal places.

(ii) Sand flows from the top half of the timer into the bottom part.
As it flows the top surfaces in both parts remain level.
At a certain time, $$98\%$$ of the sand has flowed into the bottom half of the timer.
Find $$h$$, the height of the remaining sand (in the conical part of the top of the timer).
Give your answer in $$\mbox{cm}$$, correct to $$2$$ decimal places.

(i) $$23.73\mbox{ cm}^3$$

(ii) $$0.87\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi(1.25)^3+\pi(1.25^2)(3.5)+\frac{1}{3}\pi(1.25^2)(1.5)\\&\approx23.73\mbox{ cm}^3\end{align}

(ii)

\begin{align}\frac{r}{h}=\frac{1.25}{1.5}\end{align}

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

\begin{align}r=\frac{5h}{6}\end{align}

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

\begin{align}\frac{1}{3}\pi \left(\frac{5h}{6}\right)^2(h)=(0.02)(23.73)\end{align}

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

\begin{align}h&=\sqrt\frac{(36)(3)(0.02)(23.73)}{25\pi}\\&\approx0.87\mbox{ cm}\end{align}

Video Walkthrough

## Question 8

(a) A motoring magazine collected data on cars on a particular stretch of road.
Certain details on $$800$$ cars were recorded.

(i) The ages of the $$800$$ cars were recorded. $$174$$ of them were new (less than 1 year old).
Find the $$95\%$$ confidence interval for the proportion of new cars on this road.
Give your answer correct to $$4$$ significant figures.

(ii) The data on the speeds of these $$800$$ vehicles is normally distributed with an average speed of $$87.3$$ kilometres per hour and a standard deviation of $$12$$ kilometres per hour.
What proportion of cars on this stretch of road would you expect to find travelling at over $$95$$ kilometres per hour?

(iii) The driver of a car was told that $$70\%$$ of all the speeds recorded were higher than his
speed. Find the speed at which this driver was recorded.

(i) $$0.1889<p<0.2461$$

(ii) $$0.2611$$

(iii) $$81\mbox{ km/h}$$

Solution

(i)

\begin{align}\frac{174}{800}-1.96\sqrt{\frac{\frac{174}{800}\times\frac{626}{800}}{800}}<p<\frac{174}{800}+1.96\sqrt{\frac{\frac{174}{800}\times\frac{626}{800}}{800}}\end{align}

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

\begin{align}0.1889<p<0.2461\end{align}

(ii)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{95-87.3}{12}\\&=0.64166….\end{align}

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

\begin{align}P(z\leq 0.64166…)=0.7389\end{align}

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

\begin{align}P(z\geq 0.64166…)&=1-0.7389\\&=0.2611\end{align}

(iii)

\begin{align}-0.52=\frac{x-87.3}{12}\end{align}

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

\begin{align}x&=(-0.52)(12)+87.3\\&\approx81\mbox{ km/h}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b)

(i) A road safety programme was carried out in the area using posters, signs and radio slots. After the programme the motoring magazine recorded the speeds of $$100$$ passing cars. The magazine carried out a hypothesis test, at the $$5\%$$ level of significance, to determine whether the average speed had changed.
The $$p$$-value of the test was $$0.024$$.
What can the magazine conclude based on this $$p$$-value?

(ii) The magazine found that the average speed of this sample was lower than the previously established average speed of $$87.3$$ kilometres per hour.
Find the average speed of the cars in this sample, correct to $$1$$ decimal place.

(i) We can conclude that the average speed had changed as the $$p$$-value is less than $$0.05$$.

(ii) $$84.6\mbox{ km}$$

Solution

(i) We can conclude that the average speed had changed as the $$p$$-value is less than $$0.05$$.

(ii)

\begin{align}0.024=2(1-P)\end{align}

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

\begin{align}P=0.988\end{align}

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

\begin{align}z=\pm2.26\end{align}

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

\begin{align}z=-2.26\end{align}

as we are told that the mean is lower than the previous mean.

\begin{align}-2.26=\frac{x-87.3}{12/\sqrt{100}}\end{align}

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

\begin{align}x&=87.3-\frac{2.26(12)}{10}\\&\approx84.6\mbox{ km}\end{align}

Video Walkthrough

## 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}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(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}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(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}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(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) $$29r\mbox{ m}^2$$

(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}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(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}

Video Walkthrough