Answer:
[tex]Sin \angle A =0.80[/tex]
[tex]Cos \angle A=0.60[/tex]
[tex]Sin \angle B =0.60[/tex]
[tex]Cos \angle B=0.80[/tex]
Step-by-step explanation:
Given
I will answer this question using the attached triangle
Solving (a): Sine and Cosine A
In trigonometry:
[tex]Sin \theta =\frac{Opposite}{Hypotenuse}[/tex] and
[tex]Cos \theta =\frac{Adjacent}{Hypotenuse}[/tex]
So:
[tex]Sin \angle A =\frac{BC}{BA}[/tex]
Substitute values for BC and BA
[tex]Sin \angle A =\frac{8cm}{10cm}[/tex]
[tex]Sin \angle A =\frac{8}{10}[/tex]
[tex]Sin \angle A =0.80[/tex]
[tex]Cos \angle A=\frac{AC}{BA}[/tex]
Substitute values for AC and BA
[tex]Cos \angle A=\frac{6cm}{10cm}[/tex]
[tex]Cos \angle A=\frac{6}{10}[/tex]
[tex]Cos \angle A=0.60[/tex]
Solving (b): Sine and Cosine B
In trigonometry:
[tex]Sin \theta =\frac{Opposite}{Hypotenuse}[/tex] and
[tex]Cos \theta =\frac{Adjacent}{Hypotenuse}[/tex]
So:
[tex]Sin \angle B =\frac{AC}{BA}[/tex]
Substitute values for AC and BA
[tex]Sin \angle B =\frac{6cm}{10cm}[/tex]
[tex]Sin \angle B =\frac{6}{10}[/tex]
[tex]Sin \angle B =0.60[/tex]
[tex]Cos \angle B=\frac{BC}{BA}[/tex]
Substitute values for BC and BA
[tex]Cos \angle B=\frac{8cm}{10cm}[/tex]
[tex]Cos \angle B=\frac{8}{10}[/tex]
[tex]Cos \angle B=0.80[/tex]
Using a calculator:
[tex]A = 53^{\circ}[/tex]
So:
[tex]Sin(53^{\circ}) =0.7986[/tex]
[tex]Sin(53^{\circ}) =0.80[/tex] -- approximated
[tex]Cos(53^{\circ}) = 0.6018[/tex]
[tex]Cos(53^{\circ}) = 0.60[/tex] -- approximated
[tex]B = 37^{\circ}[/tex]
So:
[tex]Sin(37^{\circ}) = 0.6018[/tex]
[tex]Sin(37^{\circ}) = 0.60[/tex] --- approximated
[tex]Cos(37^{\circ}) = 0.7986[/tex]
[tex]Cos(37^{\circ}) = 0.80[/tex] --- approximated
