Question 131335
evaluate the expression
cos [arcsin (5sqrt29/29)]

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Sketch a rt. triangle in standard position in the 1st Quadrant.
You have an angle whose sine is (5sqrt(29))/29
So the hypotenuse is 29 and the vertical side is 5sqrt(29)
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Use Pythagoras to solve for the horizontal side (hs)
(hs)^2 = 29^2 -(5sqrt(29))^2
(hs)^2 = 841 - 725
(hs) = 10.77...
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Therefore cos[angle] = 10.77/29
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Cheers,
Stan H.