# Given that tan x = 5/12, find cos x. Solve by using right-angled triangles.

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The fundamental formula in trigonometry says that:

sin^2 x+cos^2 x=1

If you divide the above formula with cos^2 x

(sin^2 x)/(cos^2 x) + 1= 1/(cos^2x)

But you know the fact that the tangent function is the ratio between sin x/cos x, so (sin^2 x)/(cos^2 x) = tg^2 x

tg^2 x + 1 = 1/(cos^2 x)

(cos^2 x)(tg^2 x + 1) = 1

cos^2 x = 1/(tg^2 x + 1)

cos x = [1/(tg^2 x + 1)]^1/2

cos x = {[1/[(5/12)^2 + 1])}^1/2

cos x = {[1/[(25/144) + 1])}^1/2

cos x = [1/(169/144)]^1/2

cos x = (144/169)^1/2

**cos x = 12/13**

1. tan x = 5/12 = opposite over adjacent.

2. The hypotenuse is a^2+B^2=c^2,

3. So, the hypotenuse is (25+144)^1/2, which is 13.

4. cos x = adjacent over hypotenuse.

5. from step one and three, we know that 12 is the length of the adjacent side and 13 is the length of the hypotenuse.

6: A/H = 12/13 = cos x.