Monthly Archives: June 2025

June 26, 2025 · 7:22 am

Trigonometry Question

I don’t know where I found this question, but it does require algebra and problem solving (as well as right trig and Pythagoras)

From a point $A$ , a lighthouse is on a bearing of $026^\circ$ T and the top of the light house is at an angle of elevation of $20.25^\circ$ . From a point $B$ , the lighthouse is on a bearing of $296^\cric$ T and the top of the lighthouse is at angle of elevation of $10.2^\circ$ . If $A$ and $B$ are 500 metres apart, find the height of the lighthouse.

Let’s draw a diagram.

Let the height of the lighthouse be $h$

We can find the angle between $A$ , the lighthouse, and $B$ by using the base triangle

The red line from $L$ is parallel to the two north lines. Hence $\theta=26^\circ+64^\circ=90^\circ$ (Alternate angles in parallel lines are congruent)

It’s a right triangle so we know

(1) $\begin{equation*}500^2=(AL)^2+(BL)^2\end{equation*}$

We are going to use the other two triangles to find $AL$ and $BL$

Substitute $AL$ and $BL$ into equation $1$

$\begin{equation*}500^2=(\frac{h}{tan(20.25)})^2+(\frac{h}{tan(10.2)})^2\end{equation}$

Solve for $h$ .

$\begin{equation*}500^2=7.35h^2+30.89h^2=38.24h^2\end{equation}$

$\begin{equation*}h^2=6538.3\end{equation}$

$\begin{equation*}h=80.9m\end{equation}$

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Filed under Algebra, Bearings, Pythagoras, Right Trigonometry, Solving Equations, Trigonometry

Tagged as hard trig, trigonometry question

June 19, 2025 · 7:18 am

Converting recurring (non-terminating) decimals to fractions

The easiest approach is to jump right in with some examples.

Example 1

Convert $0.\overline{5}$ to a fraction.

Let $x=0.\overline{5}$

(1) $\begin{equation*}x=0.\overline{5}\end{equation*}$

(2) $\begin{equation*}10x=5.\overline{5}\end{equation*}$

Subtract equation $1$ from equation $2$

$\begin{equation*}9x=5\end{equation}$

Hence $x=\frac{5}{9}$ so $0.\overline{5}=\frac{5}{9}$

Example 2

Convert $0.\overline{12}$ to a fraction.

Let $x=0.\overline{12}$

(3) $\begin{equation*}x=0.\overline{12}\end{equation*}$

(4) $\begin{equation*}100x=12.\overline{12}\end{equation*}$

Subtract equation $3$ from equation $4$ .

$\begin{equation*}99x=12\end{equation}$

$\begin{equation*}x=\frac{12}{99}=\frac{4}{33}\end{equation}$

Example 3

Convert $0.1\overline{23}$ to a fraction

Let $x=0.1\overline{23}$

(5) $\begin{equation*}x=0.1\overline{23}\end{equation*}$

(6) $\begin{equation*}10x=1.\overline{23}\end{equation*}$

(7) $\begin{equation*}1000x=123.\overline{23}\end{equation*}$

Subtract equation $6$ from equation $7$

$\begin{equation*}990x=122\end{equation}$

$\begin{equation*}x=\frac{122}{990}=\frac{61}{495}\end{equation}$

Our aim is to manipulate the recurring decimal to create two numbers each which have only the repeated digits after the decimal point.

One more example.

Example 4

Convert $3.4\overline{56}$ to a fraction

Let $x=3.4\overline{56}$

If I multiply by 10, then I will have $34.\overline{56}$ – only repeated digits after the decimal point.

If I multiply by 1000, then I will have $3456.\overline{56}$ – only repeated digits after the decimal point.

So I get,

$\begin{equation*}990x=3422\end{equation}$

$\begin{equation*}x=\frac{3422}{990}=3\frac{226}{495}\end{equation}$

You can also use your Casio classpad to do the conversion. Although I think it is easier just to do it yourself.

Let’s think about example 4,

$3.4\overline{56}=3.4+\frac{56}{1000}+\frac{56}{100000}+\frac{56}{10000000}+...$

Which is

$3.4+\frac{56}{1000\times 100^0}+\frac{56}{1000\times100^1}+\frac{56}{1000\times 100^2}...$

$3.4+\Sigma_{x=0}^\infty(\frac{56}{1000\times 100^x})$

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Filed under Arithmetic, Decimals, Fractions, Year 11 Specialist Mathematics

Tagged as recuring decimals to fractions, Year 11 Specialist Mathematics

June 10, 2025 · 1:54 am

Square Root Puzzle

Is it possible to find three numbers, $a, b, c$ , none of which is zero or a perfect square, such that
$\sqrt{a}+\sqrt{b}=\sqrt{c}$

Can You Solve These – David Wells

As $a, b$ and $c$ can’t be perfect squares, let $a=d\times e^2, b=f\times g^2$ and $c=h\times k^2$ where $d, e, f, g, h$ and $k$ are real numbers.

Hence $\sqrt{a}=e\sqrt{d}, \sqrt{b}=g\sqrt{f}$ and $\sqrt{c}=k\sqrt{h}$ .

$\begin{equation*}\sqrt{a}+\sqrt{b}=\sqrt{c}\end{equation}$

$\begin{equation*}e\sqrt{d}+g\sqrt{f}=k\sqrt{h}\end{equation}$

For the above equation to be possible $d, f$ and $h$ must simplify to the same surd. Because we are looking for one set of numbers, let $d=f=h$ .

$\begin{equation*}e\sqrt{d}+g\sqrt{d}=k\sqrt{d}\end{equation}$

$\begin{equation*}e+g=k\end{equation}$

Let’s think of some numbers that might work…

$1+2=3$ or $2+3=5$ , etc.

Let’s try $e=1, g=2,$ and $k=3$

We now have $a=d, b=4d,$ and $c=9d$

As $a$ can’t be a square number, $d$ can’t be a square number.

Try $d=2$

$\begin{equation*}\sqrt{2}+\sqrt{8}=\sqrt{18}\end{equation}$

$LHS=\sqrt{2}+2\sqrt{2}$

$LHS=3\sqrt{2}$

$LHS=\sqrt{9\times 2}$

$LHS=\sqrt{18}$

$LHS=RHS$

One set of possible numbers are $2, 8,$ and $18$ .

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Filed under Algebra, Interesting Mathematics, Puzzles

June 4, 2025 · 5:30 am

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

At 1pm, object $H$ travelling with constant velocity $\begin{pmatrix}200\\10\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}-90\\-100\end{pmatrix}$ km. At 2pm object $J$ travelling with constant velocity $\begin{pmatrix}100\\-100\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}20\\-120\end{pmatrix}$ km. Determine the minimum distance between $H$ and $J$ and when this occurs.

OT Lee Mathematics Specialist Year 11 Unit 1 and 2 Exercise 10.1 Question 6.

(1) $\begin{equation*}\mathbf{r_H}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}\end{equation*}$

(2) $\begin{equation*}\mathbf{r_J}=\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix}\end{equation*}$

$\begin{pmatrix}-80\\-20\end{pmatrix}$ is the position vector of $J$ at 1pm.

Find the relative displacement of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\mathbf{r_H}-\mathbf{r_J}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}-(\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix})\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

Find the relative velocity of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{v_J}=\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

The relative displacement is perpendicular to the relative velocity at the closest approach.

That is

(3) $\begin{equation*}\mathbf{_H}\mathbf{r_J}\cdot\mathbf{_H}\mathbf{v_J}=0\end{equation*}$

$\begin{equation*}(\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix})\cdot(\begin{pmatrix}100\\110\end{pmatrix})=0\end{equation}$