Page:VaricakRel1912.djvu/9

 In hyperbolic geometry the sum of two angles in any triangle is less than two right angles. In triangle OAB it is thus

$$\alpha_{1}+\alpha_{2}<\frac{\pi}{2},$$

If we lay off the length $$u_2$$ in the direction OC perpendicular to OA and apply under a right angle the line $$u_1$$, then we come to point D different from B. In the older mechanics those points coincide. Thus if we compose these velocities in reverse order, we obtain a resultant of same magnitude but different direction. The direction difference

can easily be represented as a function of the components.

If we introduce into the formula

the values taken from the an triangle OAB

$$\operatorname{tg}\,\alpha_{1}=\frac{\operatorname{th}\, u_{1}}{\operatorname{sh}\, u_{2}},\ \operatorname{tg}\,\alpha_{2}=\frac{\operatorname{th}\, u_{2}}{\operatorname{sh}\, u_{1}}$$

then we obtain

By (1) and (6) this goes over into

We also want to express this in a different way. The direction difference of the resultant is equal to the defect of triangle OAB. The content of a an triangle is equal to its defect. According to the known formula for the defect we can put

or