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The terminal ray of ∠A passes through the point (10,−4) . ∠A is drawn in standard position. What is the value of secA ?

 Feb 6, 2020
 #1
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sec = 1/cos

 

10, -4           use pythagorean theorem to find the hypotenuse     10^2 +(-4)^2 = h^2     h = sqrt 116

 

cos  =   adj/hyp = 10 / sqrt 116

 

sec = 1/cos =   sqrt 116/10   = 2sqrt29/10 = sqrt 29 / 5

 Feb 6, 2020
 #2
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Would the sqr only be on the 29 or for both 29/5

Guest Feb 6, 2020
 #3
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(sqrt 29) / 5

ElectricPavlov  Feb 6, 2020

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