| 5 | ||
tgα= | , oblicz długości przyprostokątnych tego trójkąta. | |
| 12 |
| 5 | ||
tga= | ||
| 12 |
| a | 5 | ||
= | |||
| b | 12 |
| 12a | ||
b= | ||
| 5 |
| 12a | ||
a2+( | )2=676 | |
| 5 |
| 144a2 | ||
a2+ | =676 | |
| 25 |
| 169a2 | |
=676 | |
| 25 |
| 12*10 | ||
b= | ||
| 5 |
| 5 | a | |||
tgα= | = | |||
| 12 | b |
| 5x | ||
tgα= | , x>0 −wspólna miara | |
| 12x |
| 262 | ||
x2= | ||
| 132 |
| 26 | ||
x= | ||
| 13 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
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| Twój nick | |