13 | ||
opisanego i wpisanego w ten trójkat jest równy | ||
4 |
R | 13 | ||
= | |||
r | 4 |
13 | ||
2R = c = | r | |
2 |
a | 2 | a | ||||
sinα = | = | * ( | ) | |||
c | 13 | r |
a + b − c | ||
r = | ||
2 |
17 | 17 | |||
a + b = | r => b = | r − a | ||
2 | 2 |
17 | 169 | |||
a2 + ( | r − a)2 = | r2 | ||
2 | 4 |
a2 | a | |||
2 | − 17 | + 30 = 0 | ||
r2 | r |
a | 13 | 13 | ||||
Podstawienie t = | −−> t ∊ (− | , | ) | |||
r | 2 | 2 |
17 − 13 | ||
t1 = | = 1 | |
4 |
30 | ||
t2 = | ||
4 |
2 | ||
sinα = | ||
13 |
π | ||
Chyba ze sinβ obliczyc z sin( | − α) | |
2 |
5 | ||
t1 = | ||
2 |
5 | ||
sinα = | , jednak co z sinusem drugiego kata? | |
13 |
2 | 12 | |||
sinα = | * 6 = | |||
13 | 13 |
2 | 5 | 5 | ||||
sinβ = | * | = | ||||
13 | 2 | 13 |