Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations by finding the sine and cosine of ∠A and ∠B using a calculator.
Answers
Answer:
angle A =sin o.98 cos=0.32
Explanation:
Sin∠A=0.80
Cos \angle A=0.60Cos∠A=0.60
Sin \angle B =0.60Sin∠B=0.60
Cos \angle B=0.80Cos∠B=0.80
Step-by-step explanation:
Given
I will answer this question using the attached triangle
Solving (a): Sine and Cosine A
In trigonometry:
Sin \theta =\frac{Opposite}{Hypotenuse}Sinθ=
Hypotenuse
Opposite
and
Cos \theta =\frac{Adjacent}{Hypotenuse}Cosθ=
Hypotenuse
Adjacent
So:
Sin \angle A =\frac{BC}{BA}Sin∠A=
BA
BC
Substitute values for BC and BA
Sin \angle A =\frac{8cm}{10cm}Sin∠A=
10cm
8cm
Sin \angle A =\frac{8}{10}Sin∠A=
10
8
Sin \angle A =0.80Sin∠A=0.80
Cos \angle A=\frac{AC}{BA}Cos∠A=
BA
AC
Substitute values for AC and BA
Cos \angle A=\frac{6cm}{10cm}Cos∠A=
10cm
6cm
Cos \angle A=\frac{6}{10}Cos∠A=
10
6
Cos \angle A=0.60Cos∠A=0.60
Solving (b): Sine and Cosine B
In trigonometry:
Sin \theta =\frac{Opposite}{Hypotenuse}Sinθ=
Hypotenuse
Opposite
and
Cos \theta =\frac{Adjacent}{Hypotenuse}Cosθ=
Hypotenuse
Adjacent
So:
Sin \angle B =\frac{AC}{BA}Sin∠B=
BA
AC
Substitute values for AC and BA
Sin \angle B =\frac{6cm}{10cm}Sin∠B=
10cm
6cm
Sin \angle B =\frac{6}{10}Sin∠B=
10
6
Sin \angle B =0.60Sin∠B=0.60
Cos \angle B=\frac{BC}{BA}Cos∠B=
BA
BC
Substitute values for BC and BA
Cos \angle B=\frac{8cm}{10cm}Cos∠B=
10cm
8cm
Cos \angle B=\frac{8}{10}Cos∠B=
10
8
Cos \angle B=0.80Cos∠B=0.80
Using a calculator:
A = 53^{\circ}A=53
∘
So:
Sin(53^{\circ}) =0.7986Sin(53
∘
)=0.7986
Sin(53^{\circ}) =0.80Sin(53
∘
)=0.80 -- approximated
Cos(53^{\circ}) = 0.6018Cos(53
∘
)=0.6018
Cos(53^{\circ}) = 0.60Cos(53
∘
)=0.60 -- approximated
B = 37^{\circ}B=37
∘
So:
Sin(37^{\circ}) = 0.6018Sin(37
∘
)=0.6018
Sin(37^{\circ}) = 0.60Sin(37
∘
)=0.60 --- approximated
Cos(37^{\circ}) = 0.7986Cos(37
∘
)=0.7986
Cos(37^{\circ}) = 0.80Cos(37
∘
)=0.80 --- approximated