Oblicz miarę kąta alfa
! Ostrosłup jest prawidłowy a h=a/pierwiastek z 6
|SD|=h=a√6
| 1 | ||
|BE|= | a | |
| 2 |
| 1 | a√3 | |||
|DE|=r= | h= | |||
| 3 | 6 |
| a√3 | ||
hs2=(a√6)2+( | )2 | |
| 2 |
| 3a2 | ||
hs2= 6a2+ | ||
| 4 |
| 27a2 | ||
hs2= | ||
| 4 |
| 3√3a | ||
hs= | ||
| 2 |
| 1 | |BE| | |||
tg( | α)= | |||
| 2 | hs |
| 1 | a | 2 | a | a√3 | ||||||
tg( | α)= | * | = | = | ||||||
| 2 | 2 | 3√3 | 3√3 | 9 |
| 1 | ||
α=2*tg( | α) | |
| 2 |
| a | ||
bo h= | ||
| √6 |
