Zadanie 1.

Kąt α jest ostry i cosα=34. Wtedy sinα jest równy

A.14

B.3–√4

C.7–√4

D.716


Zadanie 2.


Kąt α jest ostry i cosα=37. Wtedy

A.sinα=210−−√7

B.sinα=10−−√7

C.sinα=47

D.sinα=34

PLSSSSS NA JUTRO DAJE NAJ
MUSZĄ BYĆ PEŁNE OBLICZENIA BO BABKA KARZE


Odpowiedź :

1.

[tex]0^{o} < \alpha < 90^{o}\\\\cos\alpha = \frac{3}{4}\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha + (\frac{3}{4})^{2} = 1\\\\sin^{2}\alpha + \frac{9}{16} = 1\\\\sin^{2}\alpha = 1-\frac{9}{16} = \frac{16}{16}-\frac{9}{16} = \frac{7}{16}\\\\sin\alpha = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{\sqrt{16}}\\\\\boxed{sin\alpha = \frac{\sqrt{7}}{4}}[/tex]

2.

[tex]0^{o} < \alpha < 90^{o}\\\\cos\alpha = \frac{3}{7}\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha + (\frac{3}{7})^{2} = 1\\\\sin^{2}\alpha+\frac{9}{49} = 1\\\\sin^{2}\alpha = 1 - \frac{9}{49} = \frac{49}{49}-\frac{9}{49} = \frac{40}{49}\\\\sin\alpha = \sqrt{\frac{40}{49}} = \frac{\sqrt{40}}{\sqrt{49}} = \frac{\sqrt{4\cdot10}}{7}\\\\\boxed{sin\alpha = \frac{2\sqrt{10}}{7}}[/tex]