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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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<u>x₂ + x₁</u> = m         <u>y₂ + y₁</u> = m<u>
</u>    2                         2
<u>
</u>-10 = x₂
<u />9 = y₂

2=m
5=m

Plug in given numbers

<u>-10 + x</u> = 2                <u>9 + y</u><u /> = 5
      2                              2
Solve

<u>-10 + x</u> = 2*2
     2 *2

-10 + x = 4
+10        +10
         x = 14
(9 + y)/2(2) = 5 (2)
9 + y = 10
-9         -9
      y=1

Final Answer: (14,1)




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