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KonstantinChe [14]
3 years ago
10

Find sin (A-B) if sin A = 4/5 with A between 90 and 180 and if cos B = 3/5 with B between 0 and 90

Mathematics
1 answer:
quester [9]3 years ago
3 0

Answer:

sin(A-B) = 24/25

Step-by-step explanation:

The trig identity for the differnce of angles tells you ...

sin(A -B) = sin(A)cos(B) -sin(B)cos(A)

We are given that sin(A) = 4/5 in quadrant II, so cos(A) = -√(1-(4/5)^2) = -3/5.

And we are given that cos(B) = 3/5 in quadrant I, so sin(B) = 4/5.

Then ...

sin(A-B) = (4/5)(3/5) -(4/5)(-3/5) = 12/25 + 12/25 = 24/25

The desired sine is 24/25.

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Answer:

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Step-by-step explanation:

The angles both sum up to 180°. So, the equation would be: 39+(5x+1)= 180.

Step 1- Add common numbers.

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Step 2- Subtract 40 to both sides.

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Step 3- Divide both sides by 5.

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