👤
SZCZUBLEWSKAMONIKA6ON
Rozwiązane

oblicz cosa oraz tga wiedząc że kat a jest ostry i sina=0,5​

Odpowiedź :

CATTA1EYAON

[tex]\alpha\in (0; 90)\\\\sin\alpha=0.5\\sin^2\alpha+cos^2\alpha=1\\0.5^2+cos^2\alpha=1\\0.25+cos^2\alpha=1 /-0.25\\cos^2\alpha=0.75\\cos\alpha=\sqrt{0.75}=\sqrt{\frac34}=\frac{\sqrt3}2[/tex]

[tex]tg\alpha=\frac{sin\alpha}{cos\alpha}\\tg\alpha=\frac{\frac12}{\frac{\sqrt3}2}=\frac12:\frac{\sqrt3}2=\frac12*\frac2{\sqrt3}=\frac{1}{\sqrt3}=\frac{\sqrt3}3[/tex]

AGULKA1987ON

Odpowiedź:

[tex]sin \alpha = \frac{1}{2} \\ \\ {sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ ( \frac{1}{2} {)}^{2} + {cos}^{2} \alpha = 1 \\ \frac{1}{4} + {cos}^{2} \alpha = 1 \\ {cos}^{2} \alpha = \frac{3}{4} \\ cos \alpha = \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2} [/tex]

[tex]tg \alpha = \frac{sin \alpha }{cos \alpha } = \frac{ \frac{1}{2} }{ \frac{ \sqrt{3} }{2} } = \frac{1}{2} \times \frac{2}{ \sqrt{3} } = \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{3} [/tex]

On Studier: Inne Pytanie