I would like to calculate the area for a triangle such that $a^2+b^2-c^2=1$ (an almost Pythagorean triple).
I know that the triangle is non-right, so I would like to use $\text{Area}=\frac{1}{2}ab\sin C$... but I do not know how to represent $\sin C$ since I don't have any actual values.
I know about Heron's formula where $S=\frac{a+b+c}{2}$ and $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, but I feel like that gets too lengthy with our side lengths?
Edit to add: For $Area = \frac{1}{4}\sqrt{4a^2b^2-1}$ as shown by @zipirovich, can this area ever be an integer if $a,b,c$ are positive integers and $a,b >1$? Or, is this impossible?
$\endgroup$2 Answers
$\begingroup$Your equation can be rewritten as $c^2=a^2+b^2-1$. Comparing it with the Law of Cosines $c^2=a^2+b^2-2ab\cos C$, we can see that $2ab\cos C=1$ or $\cos C=\frac{1}{2ab}$. Then $$\sin C=\sqrt{1-\cos^2C}=\sqrt{1-\left(\frac{1}{2ab}\right)^2}=\sqrt{1-\frac{1}{4a^2b^2}},$$ and the area is $$\text{Area}=\frac{1}{2}ab\sin C=\frac{1}{2}ab\sqrt{1-\frac{1}{4a^2b^2}}=\frac{1}{2}\sqrt{a^2b^2-\frac{1}{4}}=\frac{1}{4}\sqrt{4a^2b^2-1}.$$ So there's no single answer because there are many such triangles, but here's an answer in the sense of having a formula for it. Going back to where we found that $\cos C=\frac{1}{2ab}$: as long as $2ab>1$, we can always find the angle $C$ and build such a triangle.
$\endgroup$ 5 $\begingroup$You can use:
\begin{equation} c^2=a^2+b^2-2ab\cos\hat C\\c^2=c^2+1-2ab\cos\hat C\\2ab\cos\hat C=1\\\cos\hat C=\frac{1}{2ab}\\ \sin\hat C=\sqrt{1-\left(\frac{1}{2ab}\right)^2} \end{equation}
So we have: $A=\frac{1}{2}ab\sqrt{1-\left(\frac{1}{2ab}\right)^2}$
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