czekam na rozwiązania
| 1 | |
2π*l = 2π*r / : 2π | |
| 4 |
| 1 | |
* 8 = r | |
| 4 |
| r | |
= sinβ | |
| l |
| 2 | 1 | |||
sin β = | = | |||
| 8 | 4 |
| 1 | 15 | |||
cos2 β = 1 − sin2 β = 1 − | = | |||
| 16 | 16 |
| √15 | ||
cos β = | ||
| 4 |
| 1 | √15 | √15 | ||||
sin α = sin 2 β = 2 sin β*cos β = 2* | * | = | ≈ 0,4841 | |||
| 4 | 4 | 8 |
Jak podał Janek r=2 , l=8
i teraz inny sposób:
α −− miara kąta rozwarcia stożka
h=√82−22= 2√15
| 1 | 1 | |||
P(ABC)= | *8*8*sinα i P(ABC)= | *4*2√15 | ||
| 2 | 2 |
| √15 | ||
32*sinα= 4√15 ⇒ sinα= | ≈0,4841 α≈29o | |
| 8 |