Monthly Archives: May 2024

May 28, 2024 · 6:52 am

Proof of the Sine Rule

As I have done a cosine rule proof, I thought I should also do a sine rule proof.

\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\textnormal{ or }\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

From the above diagram, we can find h in two ways.

sinC=\frac{h}{a}

(1)   \begin{equation*}h=asinC\end{equation*}

sinA=\frac{h}{c}

(2)   \begin{equation*}h=csinA\end{equation*}

Set equation 1 equal to equation 2

asinC=csinA

\frac{a}{sinA}=\frac{c}{sinC} or \frac{sinC}{c}=\frac{sinA}{a}

We could have put the altitude of the triangle from vertex A

Following the same process as above

sinC=\frac{h}{b}

(3)   \begin{equation*}h=bsinC\end{equation*}

sinB=\frac{h}{c}

(4)   \begin{equation*}h=csinB\end{equation*}

Set equation 3 equal to equation 4.

bsinC=csinB

\frac{b}{sinB}=\frac{c}{sinC}

Now \frac{c}{sinC}=\frac{a}{sinA} therefore

\frac{b}{sinB}=\frac{c}{sinC}=\frac{a}{sinA} or \frac{sinB}{b}=\frac{sinC}{c}=\frac{sinA}{a}

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May 26, 2024 · 1:07 pm

Quadratic Rule from a Table of Values

How do you find a quadratic rule from a table of values?

For example,

x1 2 3 4
y061424

Find the first difference

First Difference6-0=614-6=824-14=10

Find the second difference (if the second difference is a constant, then it is quadratic)

Second Difference8-6=2 10-8=2

The general equation of a quadratic is y=ax^2+bx+c

The second difference is 2a

Hence our equation is now y=x^2+bx+c

The c value is the vertical intercept (x=0). We can back track in the table

x01234
yc061424

As the first differences are 6, 8, 10, the one between 0 and 1 must be 4

0-c=4

c=-4

Our equation is now y=x^2+bx-4.

We can now use any other point to find the b value.

Let’s use the point (2, 6)

6=2^2+b(2)-4

6=4-2b-4

6=2b

b=3

The function is y=x^2+3x-4

Let’s try another one

x3456
y7173149

First differences

First difference101418

Second difference

Second Difference44

Hence 2a=4, therefore a=2

The equation is now y=2x^2+bx+c

Instead of back tracking, this time I am going to use two points and simultaneous equations.

Using points (3, 7) and (4, 17)

    \[7=2(3)^2+b(3)+c\]

(1)   \begin{equation*}3b+c=-11\end{equation*}

    \[17=2(4)^2+b(4)+c\]

(2)   \begin{equation*}4b+c=-15\end{equation*}

Equation 2 – Equation 1

b=-4

Substitute b=-4 into equation 1

3(-4)+c=-11

-12+c=-11

c=1

Hence the equation is y=2x^2-4x+1

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May 25, 2024 · 7:58 am

Proof of the cosine rule

    \[a^2=b^2+c^2-2bccos(A)\]

From the diagram above

    \[h^2=a^2-(b-x)^2\textnormal{ and }h^2=c^2-x^2\]

    \[\therefore a^2-(b-x)^2=c^2-x^2\]

    \[a^2-(b^2-2bx+x^2)=c^2-x^2\]

    \[a^2-b^2+2bx-x^2-c^2+x^2=0\]

(1)   \begin{equation*}a^2=b^2+c^2-2bx\end{equation*}

From the diagram above, we can see

    \[cos A=\frac{x}{c}\]

(2)   \begin{equation*}x=c cos A\end{equation*}

Substitute equation 2 into equation 1

    \[a^2=b^2+c^2-2bc cosA\]

It can also be handy to have the angle version

    \[cosA=\frac{b^2+c^2-a^2}{2bc}\]

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May 19, 2024 · 5:42 am

Right-Angled Trigonometry Question (very hard)

One of my year 10 students came with this question from his text book.

ICE_EM Mathematics Year 10 Third Edition

Here is my solution.

A pdf version of the solution

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May 12, 2024 · 11:50 am

Fractions to decimals

People usually know some fractions as decimals, for example

    \[\frac{1}{4}=0.25\ \textnormal{or }\frac{4}{5}=0.8\]

And denominators that are powers of ten are also easy,

    \[\frac{47}{100}=0.47\ \textnormal{or }\frac{256}{1000}=0.256\]

But what if it is something else? One that you don’t know. For example,

    \[\frac{5}{12}\ \textnormal{or }\frac{15}{37}\]

I like to do these as a long division

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May 9, 2024 · 6:08 am

Intersections of Lines and Circles

The three possibilities: no intersection, one point of intersection, or two points of intersection

Once my students have studied circle and quadratic equations, I like to introduce lines intersecting with circles. It tests their quadratic equation solving skills, and if they can use the discriminant.

Example 1

Calculate the point(s) of intersection of

    \[3x+2y-6=0\ \textnormal{and}\ (x+4)^2+(y+2)^2=9\]

Rearrange the line equation

    \[y=\frac{6-3x}{2}\]

    \[y=3-\frac{3x}{2}\]

Substitute for y into the circle equation

    \[(x+4)^2+(3-\frac{3x}{2}+2)^2=9\]

    \[(x+4)^2+(5-\frac{3x}{2})^2=9\]

    \[x^2+8x+16+25-15x+\frac{9x^2}{4}-9=0\]

    \[\frac{13x^2}{4}-7x+32=0\]

    \[13x^2-28x+128=0\]

At this point I like to check the discrimnant

    \[\Delta=b^2-4ac\]

    \[\Delta=(-28)^2-4\times13\times128\]

    \[\Delta=-5872\]

As the discriminate is less than zero, there are no points of intersection.

Example 2

Calculate the point(s) of intersection of

    \[y=2x+1\ \textnormal{and}\ (x+5)^2+(y+3)^2=16\]

    \[(x+5)^2+(2x+1+3)^2=16\]

    \[(x+5)^2+(2x+4)^2=16\]

    \[x^2+10x+25+4x^2+16x+16-16=0\]

    \[5x^2+26x+25=0\]

Check the discriminant

    \[\Delta=26^2-4\times5\times25\]

    \[\Delta=176\]

Therefore there are two points of intersection

    \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    \[x=\frac{-26\pm\sqrt{176}}{10}\]

    \[x=\frac{-26\pm4\sqrt{11}}{10}\]

    \[x=\frac{-13\pm2\sqrt{11}}{5}\]

Substitute x into y

    \[y=2(\frac{-13\pm2\sqrt{11}}{5})+1\]

    \[\textnormal{The two points are}\ (\frac{-13+2\sqrt{11}}{5},\frac{-21+4\sqrt{11}}{5})\]

    \[\textnormal{and}\ (\frac{-13-2\sqrt{11}}{5},\frac{-21-4\sqrt{11}}{5})\]

Example 3

Find the value of k so that the line and the circle intersect at one point (i.e. the line is a tangent to the circle)

    \[-kx+y-5=0\ \textnormal{and}\ (x-2)^2+(y-3)^2=\frac{36}{5}\]

    \[y=kx+5\]

    \[(x-2)^2+(kx+5-3)^2=\frac{36}{5}\]

    \[(x-2)^2+(kx+2)^2=\frac{36}{5}\]

    \[x^2-4x+4+k^2x^2+4kx+4-\frac{36}{5}=0\]

    \[(1+k^2)x^2+(4k-4)x+\frac{4}{5}=0\]

Find the discriminant (remember for one solution we want the discriminant to be zero)

    \[\Delta=b^2-4ac\]

    \[\Delta=(4k-4)^2-4\times(1+k^2)\times\frac{4}{5}\]

    \[\Delta=16k^2-32k+16-\frac{16}{5}-\frac{16k^2}{5}\]

    \[0=16k^2-32k+16-\frac{16}{5}-\frac{16k^2}{5} \[0=80k^2-160k+80-16-16k^2\]

    \[0=64k^2-160k+64\]

    \[0=32(2k^2-5k+2)\]

    \[0=2k^2-5k+2\]

    \[0=2k^2-4k-k+2\]

    \[0=2k(k-2)-1(k-2)\]

    \[0=(2k-1)(k-2)\]

    \[k=\frac{1}{2}\ \textnormal{and}\ k=2\]

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May 2, 2024 · 9:54 am

Napoleon’s Theorem

This is one of those theorems, where the name is probably not the person who discovered it. Check out the Wikipedia entry for more information on the history.

Napoleon’s theorem states that if equilateral triangles are constructed on the sides of any triangle, either all outward or all inward, the lines connecting the centres of those equilateral triangles themselves form an equilateral triangle.

I am just going to look at the outward triangles.

Construct equilateral triangles on the edges AB, BC and AC.

Find the centre (Centroid) of these triangles.

Note: The centroid is the point of intersections of the medians.

A median is the line from a vertex to the midpoint of the opposite side.

We are going to prove that triangle PQR is equilateral

I am going to add AP, AQ, QB, BR, RC and CP

As P, Q and R are the centroids, PA=PC, QA=QB, and RB=RC.

Therefore triangles AQB, BRC, and CPA are isosceles.

The angles QAB, QBA, RBC, RCB, PCA, PAC (the shaded angles in the above diagram) are 30 degrees, because the line bisects the angle (and the angle is 60 degrees because the triangles are equilateral).

Now for the maths.

    \[\textnormal {Let}\ AB=y,\ AC=x,\ BC=z\]

    \[\textnormal {Let}\ \angle{CAB}=\theta,\ \angle{ABC}=\alpha,\ \angle{BCA}=\gamma\]

    \[\therefore\angle{PAQ}=(60+\theta),\ \angle{QBR}=(60+\alpha),\]

    \[ \angle{RCP}=(60+\gamma)\]

PA is 2/3 the length of the median.

AS is the median

    \[sin(60)=\frac{AS}{x}\]

    \[AS=xsin(60)\]

    \[AS=\frac{\sqrt{3}}{2}x\]

Hence,

    \[PA=\frac{2}{3}\frac{\sqrt3}{2}x\]

    \[PA=\frac{\sqrt{3}}{3}x\]

    \[PA=\frac{x}{\sqrt{3}}\]

In the same manner,

    \[QA=\frac{y}{\sqrt{3}}\ \textnormal {and} \ BR=\frac{z}{\sqrt{3}}\]

Using the cosine rule

    \[(PQ)^2=(\frac{x}{\sqrt{3}})^2+(\frac{y}{\sqrt{3}})^2-2\frac{x}{\sqrt{3}}\frac{y}{\sqrt{3}}cos(60+\theta)\]

    \[(PQ)^2=\frac{x^2}{3}+\frac{y^2}{3}-\frac{2xy}{3}cos(60+\theta)\]

    \[\textnormal{Use a trigonometric identity to expand}\ cos(60+\theta)\]

    \[(PQ)^2=\frac{x^2}{3}+\frac{y^2}{3}-\frac{2xy}{3}(cos60cos\theta-sin60sin\theta)\]

    \[(PQ)^2=\frac{x^2}{3}+\frac{y^2}{3}-\frac{2xy}{3}(\frac{1}{2}cos\theta-\frac{\sqrt{3}}{2}sin\theta)\]

Now from the original triangle ABC

    \[cos\theta=\frac{x^2+y^2-z^2}{2xy}\ \textnormal {and}\ A=\frac{1}{2}xysin\theta\ \textnormal {where} A\ \textnormal {is the area.}\]

    \[\textnormal{Substitute into}\ (PQ)^2 \]

    \[(PQ)^2=\frac{x^2}{3}+\frac{y^2}{3}-\frac{2xy}{3}(\frac{x^2+y^2-z^2}{4xy}-\frac{\sqrt{3}A}{xy})\]

Multiply both sides by 3 and expand and simplify

    \[3(PQ)^2=x^2+y^2-2xy(\frac{x^2+y^2-z^2}{4xy}-\frac{\sqrt{3}A}{xy})\]

    \[3(PQ)^2=x^2+y^2-(\frac{x^2}{2}+\frac{y^2}{2}-\frac{z^2}{2}-2\sqrt{3}A)\]

    \[3(PQ)^2=\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}+2\sqrt{3}A\]

We are going to do the same for QR

    \[(QR)^2=\frac{y^2}{3}+\frac{z^2}{3}-\frac{2yz}{3}(cos(60+\alpha)\]

    \[(QR)^2=\frac{y^2}{3}+\frac{z^2}{3}-\frac{2xy}{3}(cos60cos\alpha-sin60sin\alpha)\]

    \[(QR)^2=\frac{y^2}{3}+\frac{z^2}{3}-\frac{2xy}{3}(\frac{1}{2}cos\alpha-\frac{\sqrt{3}}{2}sin\alpha)\]

Once again, from triangle ABC

    \[cos\alpha=\frac{y^2+z^2-x^2}{2yz}\ \textnormal {and}\ A=\frac{1}{2}zysin\alpha\ \textnormal {where} A\ \textnormal {is the area.}\]

    \[\textnormal{Substitute into}\ (QR)^2 \]

    \[(QR)^2=\frac{y^2}{3}+\frac{z^2}{3}-\frac{2yz}{3}(\frac{y^2+z^2-x^2}{4yz}-\frac{\sqrt{3}A}{yz})\]

Multiply both sides by three and expand and simplify

    \[3(QR)^2=y^2+z^2-2yz(\frac{y^2+z^2-x^2}{4yz}-\frac{\sqrt{3}A}{yz})\]

    \[3(QR)^2=y^2+z^2-(\frac{y^2}{2}+\frac{z^2}{2}-\frac{x^2}{2}-2\sqrt{3}A)\]

    \[3(QR)^2=\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}+2\sqrt{3}A\]

We can do the same for PR

    \[(PR)^2=\frac{x^2}{3}+\frac{z^2}{3}-\frac{2xz}{3}cos(60+\gamma)\]

    \[(PR)^2=\frac{x^2}{3}+\frac{z^2}{3}-\frac{2xz}{3}(cos60cos\gamma-sin60sin\gamma)\]

    \[(PR)^2=\frac{x^2}{3}+\frac{z^2}{3}-\frac{2xz}{3}(\frac{1}{2}cos\gamma-\frac{\sqrt{3}}{2}sin\gamma)\]

And from triangle ABC

    \[cos\gamma=\frac{x^2+z^2-y^2}{2xz}\ \textnormal{and}\ A=\frac{1}{2}sin\gamma\ \textnormal{where A is the area.}\]

    \[\textnormal{substitute into}\ (PR)^2\]

    \[(PR)^2=\frac{x^2}{3}+\frac{y^2}{3}-\frac{2xz}{3}(\frac{x^2+z^2-y^2}{4xz}-\frac{\sqrt{3}A}{xz})\]

Multiply both sides by three and expand and simplify

    \[3(PR)^2=\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}+2\sqrt{3}A\]

Hence, PQ=PR=RQ and triangle PQR is equilateral.

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