Dual of the formula for Euclidean triangles

Question

It’s well known that the trigonometric area formula for Euclidean triangles is

$$S=\frac{1}{2}ab\sin C\;. \tag{1.1}$$

• Is there a such formula for hyperbolic triangles?
• Is there a proof for (1.1) by using law of cosines?

Thanks.

Answer 1

You may derive a similar expression for hyperbolic triangles by combining Heron's formula $$ \tanh\frac{\Delta}{4}=\sqrt{\tanh\frac{s}{2} \tanh\frac{s-a}{2}\tanh\frac{s-b}{2}\tanh\frac{s-c}{2}},\qquad s=\frac{a+b+c}{2} \tag{1}$$ or the expression for the area in terms of the angular defect $$ \Delta = \pi-(A+B+C)\tag{2}$$ together with the laws of sines and cosines: $$ \frac{\sinh A}{\sin a}=\frac{\sinh B}{\sin b}=\frac{\sinh C}{\sin c},\qquad \cosh c=\cosh a \cosh b-\sinh a\sinh b\cos C\tag{3}$$ Given $a,b,C$, you may derive $c$, then $s$, from the law of cosines, then plug $a,b,c,s$ into $(1)$.

You can do the same in the Euclidean context.