While trying to prove Heron's formula, I stumbled on this random formula for the area of an oblique triangle: $$ \frac{(c^2 - (a^2+b^2))^2}{2a^3b^3 \cos^2 C\sin C} = A $$ where $c$ is a side opposite to angle $C$, $a$ and $b$ are two other sides, and $A$ is the area of the triangle. Is my proof for this formula correct? \begin{align} c^2 &= a^2 + b^2 - 2ab\cos C \\ c^2 - (a^2+b^2) &= -2ab\cos C \\ (c^2-(a^2+b^2))^2 &= 4a^2b^2\cos^2C \\ \frac{(c^2-(a^2+b^2))^2}{4} &= a^2b^2\cos^2C = \frac{Aa^3b^3(\cos^2C)\sin C}{2} \\ \frac{(c^2-(a^2+b^2))^2}{2a^3b^3\cos^2C\sin C} &= A \end{align}
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1I've not looked at it closely, but it doesn't look correct. For one thing, it's not dimensionally consistent. Also, explain this step: $a^2b^2cos^2C=\frac{Aa^3b^3(\cos^2C)\sin C}{2}$ (which is also dimensionally inconsistent). – Deepak Jan 05 '22 at 21:23
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1The expression actually gives you $1/A$. – bjorn93 Jan 05 '22 at 21:25
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Using dimensional analysis - you get $A = [L]^{-2}$ which does not make sense. – Chinny84 Jan 05 '22 at 21:27
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@Deepak What I did was to make this equation $\frac{1}{2}ab\sin C x = a^2b^2\cos^2C$ and then solve for $x$ – Jan 05 '22 at 21:33
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$a^2b^2\cos^2C=4A^2\cot^2 C$ – Vasili Jan 05 '22 at 21:33
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@Chinny84 what's dimensional analysis? And what's $L$? – Jan 05 '22 at 21:39
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Hi Kamal, do you have introductory knowledge of Linear Algebra? Where $x = b - a$ and $y = c - a$, and $z = x \times y$, the area is $\frac{1}{2} |z|$. – shawn_halayka Jan 05 '22 at 22:13
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@shawn_halayka what's $b$,$a$, and $c$? All I know about Linear Algebra is vectors, scalars, vector notation, determinants, and dot product. – Jan 05 '22 at 22:19
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Sorry, b a and c are the three vertices. You’ll also need to know the cross product. – shawn_halayka Jan 05 '22 at 22:46
2 Answers
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Great idea! But you need to be careful where you introduce the area $A$ using the SAS formula $$ A = \frac{1}{2} ab \sin C $$
Here's how we can repair the calculation. We have $$ \bigl( c^2 - (a^2 + b^2) \bigr)^2 = 4 a^2 b^2 \cos^2 C. $$ To avoid extra parentheses, noting that $(-u)^2 = u^2$ for all $u$, we can write: $$ ( a^2 + b^2 - c^2 )^2 = 4 a^2 b^2 \cos^2 C. $$ Now, multiply by $$ \frac{\sin C}{8ab \cos^2 C} $$ to obtain $$ A = \frac{1}{2} ab \sin C = \frac{( a^2 + b^2 - c^2 )^2 \sin C}{8ab \cos^2 C}. $$ Also, since $$ \frac{\sin C}{\cos^2 C} = \frac{1}{\cos C} \frac{\sin C}{\cos C} = \sec C \tan C, $$ we can rewrite the formula as $$ A = \frac{( a^2 + b^2 - c^2 )^2 \sec C \tan C}{8ab}. $$
Note: you have to exclude the case of right triangles so as to avoid division by $0$. This formula applies to all other triangles though!
Sammy Black
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Thanks for your help! I think what I can say is $c^2$ is not equal to $a^2+b^2$. – Jan 13 '22 at 01:43
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Since $\frac12ab\sin C=A$, you accidentally multiplied $a^2b^2\cos^2C$ by $A^2$ instead of $1$ in one of your $=$ signs. We can correct this as$$\frac{(a^2+b^2-c^2)^2}{4}=a^2b^2\cos^2C\frac{2A}{ab\sin C}=\frac{2Aab\cos^2C}{\sin C}\implies A=\frac{(a^2+b^2-c^2)^2\sin C}{8ab\cos^2C}$$or infinitely many alternatives, including$$\frac{(a^2+b^2-c^2)^2}{4}=a^2b^2\cos^2C\frac{ab\sin C}{2A}\implies A=\frac{2a^3b^3\cos^2C\sin C}{(a^2+b^2-c^2)^2}.$$The first of these is @SammyBlack's result. They are both dimensionally consistent in that they have length dimension $2$, as an area should, be it $2\times2-2=2$ or $3\times2-2\times2=2$ (because e.g. $a^n$ has length dimension $n$). Your formula, by contrast, has length dimension $2\times2-3\times2=-2$.
J.G.
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Quick question, so where do you get the twos and the three from (when you explained dimensionally consistency)? – Jan 13 '22 at 01:42
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@KamalSaleh Combine $\dim a^n=n$ etc. with $\dim\prod_ix_i^{n_i}=\sum_in_i\dim x_i$. – J.G. Jan 13 '22 at 07:33
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