G3.2 It is known that $\sqrt{\frac{50+120+130}{2} \times(150-50) \times(150-120) \times(150-130)}=\frac{50 \times 130 \times k}{2}$. If $t=\frac{k}{\sqrt{1-k^{2}}}$, find the value of $t$.
The question is equivalent to: given a triangle with sides $50,120,130$, find its area. $\cos C=\frac{50^{2}+130^{2}-120^{2}}{2 \cdot 50 \cdot 130}=\frac{5}{13}$
Using the formula $\frac{1}{2} a b \sin C=\frac{50 \times 130 \times k}{2}, k=\sin C=\sqrt{1-\cos ^{2} C}=\frac{12}{13}$ $t=\frac{k}{\sqrt{1-k^{2}}}=\frac{\sin C}{\cos C}=\tan C=\frac{12}{5}$