Let ABC be a triangle with side lengths AB = 5, BC = 6, and AC = 9. What is the area of the triangle with side lengths \tan A, \tan B, and \tan C?
What is the area of the triangle with side lengths tan A, tan B, and tan C?
\(f_{AC}(x)=\sqrt{9^2-x^2}\\ f_{BC}(x)=\sqrt{6^2-x^2+10x-5^2}\\ 81-x^2=36-x^2+10x-25\\ 10x=81-36+25\\ x_C=\dfrac{81-36+25}{10}\\ x_C=7\\ y_C=\sqrt{9^2-7^2}\\ C\ (7,5.6569)\)
\(\angle B=180^\circ -atan\ \frac{y_c}{x_C-x_B}=180^\circ-atan\ \frac{5.6569}{7-5}\\ \angle B=109.4712^\circ\\ \color{red}tan\ \angle B=-2.8284\\ The\ sides\ of\ a\ real\ triangle\ are\ greater\ than\ zero. \)
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