| 7 | ||
tg(90o − x) = | ||
| 24 |
| sin(90o−x) | cosx | |||
zrobiłam tak tg(90o − x) = | = | = | ||
| cos(90o−x) | sinx |
| 7 | 24 | |||
= ctgx = | → tgx = | |||
| 24 | 7 |
| 24 | 7 | |||
w odpowiedziach mam, że sinx= | , a cosx= | |||
| 25 | 25 |
α+β=90o ⇒ (90o−α)=β
| 7 | 24 | 7 | 24 | |||||
tgβ= | to sin α= | , cosα= | tgα= | |||||
| 24 | 25 | 25 | 7 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
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