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patriot [66]
3 years ago
15

If a and b are two angles in standard position in Quadrant I, find cos(a+b) for the given function values. sin a=4/5 and cos b=5

/13
Mathematics
2 answers:
soldier1979 [14.2K]3 years ago
5 0

Answer:

cos(a+b)≈0.507

Step-by-step explanation:

Hi there, in Quadrant I both sine and cosine functions are positive.

This cos (a+b) being a Classical Trigonometrical Identity, the product of the sum between two cosines equals the product of two cosines minus the product of two sines.

But we need to find the value of a and b, so that we can go on. Calculate the arccosine and arcsine function respectively is mandatory

cos(a+b)=cosa*cosb-sena*senb\\ cos(a+b)=cosa*\frac{5}{13} -\frac{4}{5}*senb\\ a=arcsin(\frac{5}{13})\\a=22.61\\b=arccos(\frac{4}{5})\\b=36.86\\ 5/13=0.38\\4/5=0.8\\cos(a+b)=cos(22.61)*cos(36.86) -sen(22.61)*sen(36.86)\\ cos (a+b)=0.507

vladimir1956 [14]3 years ago
4 0
First let's find the angles a and b.

 We have then:
 sin a = 4/5
 a = Asin (4/5)
 a = 53.13 degrees.

 cos b = 5/13
 b = Acos5 / 13
 b = 67.38 degrees.

 We now calculate cos (a + b). To do this, we replace the previously found values:
 cos ((53.13) + (67.38)) = - 0.507688738
 Answer: 
 -0.507688738
 Note: there is another way to solve the problem using trigonometric identities.
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Find the value of x then tell whether the side lengths form a Pythagorean triple 6 , 10 , x
masya89 [10]

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Explanation:

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10x + 4y = 24 6x +12y = 48 What is the value of y? * <br>A. 7 <br>B. 7/2 <br>C. 1 <br>D. -7/2​
ziro4ka [17]

Answer:

B

Step-by-step explanation:

1) lets use elimination to solve this:

multiply 10x+4y=24 with 3 in order to get "4y" into "12y"

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30x+12y=72

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6x+12y=48

---------------------

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6(1)+12y=48

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3 years ago
Is a triangle with the angles 60 degrees, 40 degrees, and 80 degrees unique?
oksano4ka [1.4K]

Hello,

No a triangle with these angles are not unique. This is because every triangle is supposed to have 180 degrees, just like this one.

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