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Artyom0805 [142]
3 years ago
13

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
Mathematics
2 answers:
marusya05 [52]3 years ago
8 0

Answer:

Sin \angle A =0.80

Cos \angle A=0.60

Sin \angle B =0.60

Cos \angle 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} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle A =\frac{BC}{BA}

Substitute values for BC and BA

Sin \angle A =\frac{8cm}{10cm}

Sin \angle A =\frac{8}{10}

Sin \angle A =0.80

Cos \angle A=\frac{AC}{BA}

Substitute values for AC and BA

Cos \angle A=\frac{6cm}{10cm}

Cos \angle A=\frac{6}{10}

Cos \angle A=0.60

Solving (b): Sine and Cosine B

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle B =\frac{AC}{BA}

Substitute values for AC and BA

Sin \angle B =\frac{6cm}{10cm}

Sin \angle B =\frac{6}{10}

Sin \angle B =0.60

Cos \angle B=\frac{BC}{BA}

Substitute values for BC and BA

Cos \angle B=\frac{8cm}{10cm}

Cos \angle B=\frac{8}{10}

Cos \angle B=0.80

Using a calculator:

A = 53^{\circ}

So:

Sin(53^{\circ}) =0.7986

Sin(53^{\circ}) =0.80 -- approximated

Cos(53^{\circ}) = 0.6018

Cos(53^{\circ}) = 0.60 -- approximated

B = 37^{\circ}

So:

Sin(37^{\circ}) = 0.6018

Sin(37^{\circ}) = 0.60 --- approximated

Cos(37^{\circ}) = 0.7986

Cos(37^{\circ}) = 0.80 --- approximated

Nat2105 [25]3 years ago
3 0

Answer:

Lengths

AB = 13

BC = 5

AC = 12

A Sin = 0.38 A Cos = 0.92

B sin = 0.92 B Cos = 0.38

Also there is a image of the answer

Step-by-step explanation: for Plato/edmentum

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