Category Archives: Right Trigonometry

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an n-sided regular polygon if you know the side length, l.

An octagon for a visual reference

Find the h of the triangle in terms of l and theta.

tan(\theta)=\frac{\frac{l}{2}}{h}

h=\frac{l}{2tan(\theta)}

Remember the area of a triangle is A=\frac{1}{2}bh

Hence, A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}

And \theta=\frac{360}{2n}=\frac{180}{n}

Therefore A=\frac{l^2}{4tan(\frac{180}{n})}

There are n triangles in an n-sided polygon

(1)   \begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}

Find the area of a hexagon with side length 10cm.
A=\frac{6\times10^2}{4tan(\frac{180}{6})}
A=\frac{600}{4(\frac{1}{\sqrt{3}})}
A=150\sqrt{3} cm^2

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of a and theta.

tan(\theta)=\frac{\frac{l}{2}}{a}

l=2atan(\theta)

A=\frac{1}{2}bh

A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)

And \theta=\frac{180}{n}

Hence for an n-sided polygon

(2)   \begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}

Find the area of a regular pentagon with apothem 4.5cm
A=5\times 4.5^2tan(\frac{180}{5})
A=73.56cm^2

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle (R in the diagram above)

Remember the area of triangle formula

A=\frac{1}{2}absin(\theta)

A=\frac{1}{2}R^2sin(\theta)

\theta=\frac{360}{n}

Hence, A=\frac{1}{2}R^2sin(\frac{360}{n})

Hence, for an n-sided polygon

(3)   \begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}

Find the area of a regular octagon inscribed in a circle of radius 10cm.
A=\frac{8\times 10^2sin(45)}{2}
A=200\sqrt{2}cm^2

Leave a Comment

Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Related Rates Question

A boat is moving towards the beach line at 3 metres/minute. On the boat is a rotating light, revolving at 4 revolutions/minute clockwise, as observed from the beach. There is a long straight wall on the beach line, as the boat approaches the beach, the light moves along the wall. Let x equal the displacement of the light from the point O on the wall, which faces the boat directly. See the diagram below.
Determine the velocity, in metres/minute, of the light when x=5 metres, and the distance of the boat from the beach D is 12 metres.

Mathematics Specialist Semester 2 Exam 2018


The light is rotating at 4 revolutions/minute, which means

    \begin{equation*}\frac{d\theta}{dt}=4\times\pi\end{equation}

We want to find \frac{dx}{dt} and we know \frac{d\theta}{dt} and \frac{dD}{dt}.

We need to find an equation connecting x, \theta, and D.

    \begin{equation*}tan (\theta)=\frac{x}{D}\end{equation}

Differentiate (implicitly) with respect to time.

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

Now we know x=5, and D=12, using pythagoras we can calulate the hypotenuse.


h=\sqrt{12^2+3^2}=13
sec(\theta)=\frac{13}{12}

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

    \begin{equation*}(\frac{13}{12})^2\times 4\pi=\frac{\frac{dx}{dt}12-3\times 5}{12^2}\end{equation}

    \begin{equation*}169\times4\pi=12\frac{dx}{dt}-15\end{equation}

    \begin{equation*}676\pi+15=12\frac{dx}{dt}\end{equation}

    \begin{equation*}\frac{dx}{dt}=178.2\end{equation}

The velocity of the light is 178.2 m/minute.

Leave a Comment

Filed under Differentiation, Implicit, Pythagoras, Right Trigonometry, Trigonometry, Uncategorized

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}

Leave a Comment

Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry

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

    \[cosA=\frac{b^2+c^2-a^2}{2bc}\]

1 Comment

Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry, Uncategorized

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

Leave a Comment

Filed under Right Trigonometry, Trigonometry, Uncategorized