Category Archives: Right Trigonometry

September 30, 2025 · 3:54 am

Geometry Problem

*Geometry Snacks* by Ed Southall and Vincent Pantaloni

I started by trisecting another side of the triangle

This makes it clearer that the two lines are parallel

Which means the two angles labelled above are corresponding and therefore congruent.

Let the side length be $x$ .

The area of the equilateral triangle is

(1) $\begin{equation*}A=\frac{1}{2}x^2 sin(60)=\frac{\sqrt{3}x^2}{4}\end{equation*}$

$\begin{equation*}cos(60)=\frac{y}{\frac{2x}{3}}\end{equation}$

$\begin{equation*}y=\frac{x}{3}\end{equation}$

Area of right triangle

(2) $\begin{equation*}A=(\frac{1}{2})(\frac{2x}{3})(\frac{x}{3})sin(60)=\frac{\sqrt{3}x^2}{18}\end{equation*}$

The fraction of the area is

$\begin{equation*}=\frac{\frac{\sqrt{3}x^2}{18}}{\frac{\sqrt{3}x^2}{4}}=\frac{2}{9}\end{equation}$

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Filed under Algebra, Area, Area of Triangles (Sine), Finding an area, Geometry, Puzzles, Right Trigonometry, Simplifying fractions, Trigonometry

Tagged as geometry problem, geometry snacks

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

April 7, 2025 · 9:09 am

Vector and Scalar Projection

The vector projection (vector resolution or vector component) of $\mathbf{a}$ onto a non-zero vector $\mathbf{b}$ is splitting $\mathbf{a}$ into two vectors, one is parallel to $\mathbf{b}$ (the vector projection) and one perpendicular to $\mathbf{b}$

In the above diagram $\mathbf{a_1}$ is the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ and $\mathbf{a_2}$ is perpendicular to $\mathbf{b}$ .

How do we find $\mathbf{a_1}$ and $\mathbf{a_2}$ ?

Using right trigonometry,

$cos(\theta)=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

Remember the scalar product (dot product) of vectors is

(1) $\begin{equation*}\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|cos(\theta)\end{equation*}$

Hence $cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

$\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

and, $|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\times|\mathbf{a}|$

$|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$

This is the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$

To find the vector projection we need to multiply by $\mathbf{\hat{b}}$ , that is find a vector with the same magnitude as $\mathbf{a_1}$ in the direction of $\mathbf{b}$ .

The vector projection is

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \mathbf{\hat{b}}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \frac{\mathbf{b}}{|\mathbf{b}|}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{(|\mathbf{b}|)^2}\times \mathbf{b}$

Now for $\mathbf{a_2}$ , we know $\mathbf{a}=\mathbf{a_1}+\mathbf{a_2}$

Hence, $\mathbf{a_2}=\mathbf{a}-\mathbf{a_1}$

Example

From Cambridge Year 11 Specialist Mathematics (Chapter 3)

(a) $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$

$=\frac{\begin{pmatrix}4\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\end{pmatrix}}{2}\times(\mathbf{i}-\mathbf{j})$

$=\frac{3}{2}{(\mathbf{i}-\mathbf{j})$

(b) $4\mathbf{i}+\mathbf{j}-\frac{3}{2}{(\mathbf{i}-\mathbf{j})=\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

(c)

The shortest distance (green vector) is the vector component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ , i.e. $\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

$|\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}|=\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}$

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Filed under Right Trigonometry, Trigonometry, Vector Projection, Vectors, Year 11 Specialist Mathematics

Tagged as dot product, scalar product, scalar projection, vector projection

April 3, 2025 · 7:28 am

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, $l$ .

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

A=\frac{6\times10^2}{4tan(\frac{180}{6})}

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

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

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

Tagged as area of regular polygons, area of triangles, trig

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.

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

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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Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry

Tagged as proof of sine rule, sine rule, sine rule proof

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

Tagged as cosine rule, proof of the cosine rule

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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Filed under Right Trigonometry, Trigonometry, Uncategorized

Tagged as 3D trigonometry, right-angled trigonometry, year 10

Category Archives: Right Trigonometry

Geometry Problem

Trigonometry Question

Vector and Scalar Projection

Example

Area of Regular Polygons

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

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

Finding the area of a regular polygon given the circumradius

Proof of the Sine Rule

Proof of the cosine rule

Right-Angled Trigonometry Question (very hard)

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