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Question 2 Given $\csc \theta = \frac{17}{15}$, determine the values for "a" and "b" in $\cos \theta = \frac{a}{b}$ a = b =

          Question 2
Given $\csc \theta = \frac{17}{15}$, determine the values for "a" and "b" in $\cos \theta = \frac{a}{b}$
a = 
b =
        
Question 2
Given cscθ = (17)/(15), determine the values for "a" and "b" in cosθ = (a)/(b)
a = 
b =

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Question 2 Given csc heta =(17)/(15), determine the values for "a" and "b" in cos heta =(a)/(b) a= b= Now solve for x to the nearest degree: x= Question 2 17 determine the values for "a" and "b" in cos e : 15 a b Given csc 0 = b =
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Transcript

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00:02 I want to find the tangent of a plus b.
00:08 That formula says to take the tangent of the first angle, add to it the tangent of the second angle, divide that by the difference 1 minus tan a, tan b.
00:22 So you can see from the formula i need the tangent of each of these angles.
00:29 A is in quadrant 2, and i have the sign, so i have 3.
00:35 The opposite side and the hypotenuse.
00:38 I'm going to solve for the adjacent side using the pythagorean theorem.
00:45 So a squared is going to equal 45.
00:49 A is the square root of 45, which simplifies three radical five.
00:57 So the tangent of a is going to be opposite, which is two over adjacent, three rad five, and it's negative in quadrant two.
01:09 Let's do the same thing with b.
01:12 B is in quadrant 1.
01:15 Cosine is adjacent over hypotenuse.
01:19 So a squared plus 2 squared equals 3 squared.
01:26 A squared is 5.
01:28 A is the square root of 5.
01:31 Tangent of b is opposite over adjacent, positive in quadrant 1...
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