Question

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

Answer 1

Lengths

AB = 13

BC = 5

AC = 12

A Sin = 0.38 A Cos = 0.92

B sin = 0.92 B Cos = 0.38

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Step-by-step explanation: for Plato/edmentum

Answer 2

`Sin \angle A =0.80`

`Cos \angle A=0.60`

`Sin \angle B =0.60`

`Cos \angle B=0.80`

Explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

`Sin \theta =(Opposite)/(Hypotenuse)` and

`Cos \theta =(Adjacent)/(Hypotenuse)`

So:

`Sin \angle A =(BC)/(BA)`

Substitute values for BC and BA

`Sin \angle A =(8cm)/(10cm)`

`Sin \angle A =(8)/(10)`

`Sin \angle A =0.80`

`Cos \angle A=(AC)/(BA)`

Substitute values for AC and BA

`Cos \angle A=(6cm)/(10cm)`

`Cos \angle A=(6)/(10)`

`Cos \angle A=0.60`

Solving (b): Sine and Cosine B

In trigonometry:

`Sin \theta =(Opposite)/(Hypotenuse)` and

`Cos \theta =(Adjacent)/(Hypotenuse)`

So:

`Sin \angle B =(AC)/(BA)`

Substitute values for AC and BA

`Sin \angle B =(6cm)/(10cm)`

`Sin \angle B =(6)/(10)`

`Sin \angle B =0.60`

`Cos \angle B=(BC)/(BA)`

Substitute values for BC and BA

`Cos \angle B=(8cm)/(10cm)`

`Cos \angle B=(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