Category Archives: Right Trigonometry

October 28, 2024 · 6:51 am

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

Tagged as ,

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}

Leave a Comment

Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry

Tagged as , ,

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

    \[cosA=\frac{b^2+c^2-a^2}{2bc}\]

1 Comment

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

Tagged as ,

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

Leave a Comment

Filed under Right Trigonometry, Trigonometry, Uncategorized

Tagged as , ,