As I have done a cosine rule proof, I thought I should also do a sine rule proof.
From the above diagram, we can find 
in two ways.
Set equation 1 equal to equation 2
or 
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.
Now 
therefore
or 
Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry
![\[a^2=b^2+c^2-2bccos(A)\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-99c9dacc9dda0545a30e7c65edb0e5bb_l3.png "Rendered by QuickLaTeX.com")
From the diagram above
![\[h^2=a^2-(b-x)^2\textnormal{ and }h^2=c^2-x^2\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-54637408ff04bc1bf1ccd1d6076188a9_l3.png "Rendered by QuickLaTeX.com")
![\[\therefore a^2-(b-x)^2=c^2-x^2\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-7f1d6f4b939731176c1746c183d4bd58_l3.png "Rendered by QuickLaTeX.com")
![\[a^2-(b^2-2bx+x^2)=c^2-x^2\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-4c833336f00a15c12594cdd438562025_l3.png "Rendered by QuickLaTeX.com")
![\[a^2-b^2+2bx-x^2-c^2+x^2=0\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-5ab8b2187aa46934c6cee65a6eee8f64_l3.png "Rendered by QuickLaTeX.com")
(1) 
From the diagram above, we can see
![\[cos A=\frac{x}{c}\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-0ef2b64172bd3cefda68b0461f53508b_l3.png "Rendered by QuickLaTeX.com")
(2) 
Substitute equation 2 into equation 1
![\[a^2=b^2+c^2-2bc cosA\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-57236d6162f2cdcefc50d516d4a1664c_l3.png "Rendered by QuickLaTeX.com")
It can also be handy to have the angle version
![\[cosA=\frac{b^2+c^2-a^2}{2bc}\]](https://www.racquelsanderson.com/wp-content/ql-cache/quicklatex.com-3e6c8a9430c138c2b9a449eead64fc03_l3.png "Rendered by QuickLaTeX.com")
Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry, Uncategorized
One of my year 10 students came with this question from his text book.
Here is my solution.
A pdf version of the solution
Filed under Right Trigonometry, Trigonometry, Uncategorized
This is question 5 from the UK Maths Trust Senior Challenge October 2023.
I have tackled this in three ways; using non-right trig to find the area, Heron’s Law, and the Shoelace Formula.
Use the area of a triangle formula
Use the cosine rule to find cosθ.
Once we have cosθ, we can find sinθ.
Hence the area is,
Use Heron’s law.
Heron’s law is a way of calculating area of a triangle from the lengths of the three sides of the triangle.
This is my preferred method – simple and direct.
Shoelace formula (Gauss’s Area formula)
We need to allocate each of the vertices a co-ordinate.
The co-ordinates are listed in an anti-clockwise direction.
This is probably a bit over the top, but once you get the hang of it, it’s very easy.
