11,17,22 lang po yung sagutan niyo po
pasagot po salamat .
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Step-by-step explanation:
11. α = sin⁻¹ ([tex]\frac{1}{2}[/tex]) = 30 or 30° (using the special triangle, 1/2 is sin30. so the angle is 30.)
12. Θ = cos⁻¹ ([tex]\frac{1}{2}[/tex]) = 60°
13. β = tan⁻¹ [tex](\frac{\sqrt{3}}{3})[/tex] the same as tan⁻¹ (1/√3), by multiplying √3 for both numerator and denominator.
β = tan⁻¹ [tex](\frac{\sqrt{3}}{3})[/tex] =30°
14. α = sin⁻¹ ([tex]\frac{\sqrt{3}}{2}[/tex]) = 60°
15. Θ = cos⁻¹ ([tex]\frac{\sqrt{2}}{2}[/tex]) = 45°
16. α = sin⁻¹ ([tex]\frac{\sqrt{2}}{2}[/tex]) = 45°
17. tan α = √3 = [tex]\frac{\sqrt{3}}{1}[/tex]
α = tan⁻¹ [tex](\frac{\sqrt{3}}{1})[/tex] = 60°
18. cot β = √3 = [tex]\frac{\sqrt{3}}{1}[/tex] cot β = [tex]\frac{1}{tan\beta}[/tex]
so: 1 / tan β = √3
tan β = [tex]\frac{1}{\sqrt{3}}[/tex]
β = tan⁻¹ [tex](\frac{1}{\sqrt{3}})[/tex] = 30°
19. cot α = 1 or [tex]\frac{1}{1}[/tex] or 1/tan α = 1 or tan α =1
checking at my attached drawings for special triangles, the answer is 45°
or α = cot⁻¹ 1 or tan⁻¹ 1 = 45°
20. sec α = √2 == [tex]\frac{\sqrt{2}}{1}[/tex] == 1 / cos α
cos α = 1 / √2
α = cos⁻¹ (1/ √2) = 45°
21. csc α = √2 == [tex]\frac{\sqrt{2}}{1}[/tex] == 1 / sin α
sin α = 1 / √2
α = sin⁻¹ (1/ √2) = 45°
22. sec α = [tex]\frac{2\sqrt{3}}{3}[/tex] == 1 / cos α
also: [tex]\frac{2\sqrt{3}}{3}[/tex] = [tex]\frac{2}{\sqrt{3}}[/tex] (just multiply the numerator and denominator with √3)
cos α = √3 / 2
α = cos⁻¹ (√3 / 2) = 30°