Zamień miarę łukową...- Zadanie 6.6.: Prosto do matury 3. Zakres podstawowy i rozszerzony

Rozwiązanie

$a) α_{1} = \frac{x _{1} \cdot 180 ^{\circ}}{π} = \frac{\frac{π}{6} \cdot 180 ^{\circ}}{π} = \frac{180 ^{\circ}}{6} = 30^{\circ}$

$α_{2} = \frac{x _{2} \cdot 180 ^{\circ}}{π} = \frac{\frac{3}{4} π \cdot 180 ^{\circ}}{π} = \frac{3}{4} \cdot 180^{\circ} = 3 \cdot 45^{\circ} = 135^{\circ}$

$α_{3} = \frac{x _{3} \cdot 180 ^{\circ}}{π} = \frac{\frac{23}{18} π \cdot 180 ^{\circ}}{π} = 23 \cdot 10^{\circ} = 230^{\circ}$

$b) α_{1} = \frac{x _{1} \cdot 180 ^{\circ}}{π} = \frac{\frac{π}{9} \cdot 180 ^{\circ}}{π} = \frac{180 ^{\circ}}{9} = 20^{\circ}$

$α_{2} = \frac{x _{2} \cdot 180 ^{\circ}}{π} = \frac{\frac{2 π}{5} \cdot 180 ^{\circ}}{π} = \frac{2}{5} \cdot 180^{\circ} = 2 \cdot 36^{\circ} = 72^{\circ}$

$α_{3} = \frac{x _{3} \cdot 180 ^{\circ}}{π} = \frac{\frac{10 π}{3} \cdot 180 ^{\circ}}{π} = \frac{10}{3} \cdot 180^{\circ} = 10 \cdot 60^{\circ} = 600^{\circ}$

$c) α_{1} = \frac{x _{1} \cdot 180 ^{\circ}}{π} = \frac{2 \cdot 180 ^{\circ}}{π} = (\frac{360}{π})^{\circ}$

$α_{2} = \frac{x _{2} \cdot 180 ^{\circ}}{π} = \frac{1 , 5 \cdot 180 ^{\circ}}{π} = (\frac{270}{π})^{\circ}$

$α_{3} = \frac{x _{3} \cdot 180 ^{\circ}}{π} = \frac{0 , 7 \cdot 180 ^{\circ}}{π} = (\frac{126}{π})^{\circ}$

$d) α_{1} = \frac{x _{1} \cdot 180 ^{\circ}}{π} = \frac{102 \cdot 180 ^{\circ}}{π} = (\frac{18 360}{π})^{\circ}$

$α_{2} = \frac{x _{2} \cdot 180 ^{\circ}}{π} = \frac{1 , 25 \cdot 180 ^{\circ}}{π} = (\frac{225}{π})^{\circ}$

$α_{3} = \frac{x _{3} \cdot 180 ^{\circ}}{π} = \frac{7 , 5 \cdot 180 ^{\circ}}{π} = (\frac{1350}{π})^{\circ}$

Natalia Wodka

Nauczycielka matematyki

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