I have an Equilateral triangle with unknown side $a$. The next thing I do is to make a random point inside the triangle P. The distance $|AP|=3 cm, |BP|=4 cm, |CP|=5 cm.$ What is the area of the triangle?
I have seen this problem posted here before and solved, but done without trigonometry. How would you do this problem with trigonometry?
Here my point is $D$ instead of $P$, $s$ is side of triangle: $|AD|=3,|BD|=4,|CD|=5$ Using cosine rule: $$\cos \angle ADB=\frac{9+16-s^2}{24}\\ \cos \angle BDC=\frac{16+25-s^2}{40}\\ \cos \angle CDA=\frac{25+9-s^2}{30}\\ $$ Also $\angle ADB+\angle BDC+\angle CDA=2\pi$
So, $cos \angle CDA=\cos (2\pi-(\angle ADB+\angle BDC))=\cos(\angle ADB+\angle BDC)$
$$\begin{align} cos \angle CDA&=\cos(\angle ADB+\angle BDC)\\ cos \angle CDA&=\cos(\angle ADB)\cos(\angle BDC)-\sin(\angle ADB)\sin(\angle BDC)\\ \end{align}$$ Now we need to solve: $$\frac{25+9-s^2}{30}=\frac{9+16-s^2}{24}\frac{16+25-s^2}{40}-\sqrt{1-\left(\frac{9+16-s^2}{24}\right)^2}\sqrt{1-\left(\frac{16+25-s^2}{40}\right)^2} $$ Or: $$\left(\frac{25+9-s^2}{30}-\frac{9+16-s^2}{24}\frac{16+25-s^2}{40}\right)^2=\left(1-\left(\frac{9+16-s^2}{24}\right)^2\right)\left(1-\left(\frac{16+25-s^2}{40}\right)^2\right)$$ Or: $$\frac{s^8}{921600}-17\frac{s^6}{230400}+103\frac{s^4}{92160}+119\frac{s^2}{25600}+441\frac1{102400}=\frac{s^8}{921600}-11\frac{s^6}{76800}+47\frac{s^4}{10240}-2017\frac{s^2}{230400}+441\frac1{102400}$$ Or: $$\frac{s^6}{14400}-\frac{s^4}{288}+\frac{193s^2}{14400}=0$$ As $s\neq0\Longleftrightarrow s^2\neq0$: $$s^4-50s^2+193=0$$ So: $$(s^-25)^2=432$$ Which gives: $$s=\pm_1\sqrt{25\pm_212\sqrt3}$$ Since $s>0$: $s=\sqrt{25\pm12\sqrt3}\approx 2.0531,6.7664$$
