What is cos x if tan x=6/11?Solve using right triangle
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We have that tan x = 6/11.
tan x is equal to the opposite side/ adjacent side for the angle in a right triangle.
Using the Pythagorean theorem we calculate the length of the hypotenuse as sqrt ( 6^2 + 11^2) = sqrt ( 36 + 121) = sqrt 157.
Now, cos x = adjacent side/ hypotenuse
=> 11 / sqrt 157.
cos x = 11 / sqrt 157
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We'll write the fundamental formula in trigonometry:
(sin x)^2 + (cos x)^2 = 1
If you divide the above formula with (cos x)^2
(sin x)^2/(cos x)^2 + 1= 1/(cos x)^2
But the tangent function is the ratio between sin x/cos x, so (sin x)^2/(cos x)^2 = (tan x)^2
(tan x)^2+ 1 = 1/(cos x)^2
(cos x)^2[(tan x)^2+ 1] = 1
(cos x)^2 = 1/[(tan x)^2+ 1]
cos x = sqrt1/[(tan x)^2+ 1]
cos x = sqrt[1/[(6/11)^2 + 1]
cos x = sqrt[1/[(36/121) + 1]
cos x = sqrt[1/(36+121)/121]
cos x = sqrt (121/157)
cos x = 11/sqrt 157
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