# right angle trianglei have to calculate cosine of angle using tanx=10/24 and right angle triangle. how come?

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For the right angle triangle, let x be one of the other angles.

Then we know that:

cosx = adjacent side/ hypotenuse.

But we are given that tanx = 10/24

But we know that: tanx = oppsite side / adjacent.

==> Then , we will assume that the adjacent side= 24, and the opposite side = 10

Now since we have both sides, we will calcultae the hypotenuse.

==> hypotenuse^2 = 10^2 + 24^2

==> hypotenuse^2 = 100+ 576 = 676

==> hypotenuse = sqrt676= 26

Now we will calculate the cosines.

==> cosx = adjacent/ hypotenuse = 24/26 = 12/13

**==> cosx = 12/13**

To calculate cosine, using adjacent dided by hypotenuse.

Tangent is opposite over adjacent.

opposite =10

adjacent = 24

So you are missing hypotenuse. In a right triangle, c is the hypotenuse. Pythagorean theorem states:

a squared + b squared = c squared

10 squared + 24 squared = c squared

100 + 576 = c squared

676 = c squared

square root of 676 = c

c = 26

cosign= 24/26

cosine x= .92

You have tan x = 10/24

(tan x)^2 = (10/24)^2

=> (sin x)^2 / (cos x)^2 = (10/ 24)^2

=> [1 - (cos x)^2] / (cos x)^2 = (10/ 24)^2

=> [1 - (cos x)^2] / (cos x)^2 = 25 / 144

=> 144*[1 - (cos x)^2] = 25* (cos x)^2

=> 144 - 144*( cos x)^2 = 25* (cos x)^2

=> 144 = (144 + 25)(cos x)^2

=> cos x = sqrt (144/ 169)

**=> cos x = 12/13**

First, we'll divide by 2 the numerator and the denominator of the ratio 10/24.

tan x = 5/12

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

If you divide the above formula by (cos x)^2:

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

tan x = sin x/cos x, so(sin x)^2/(cos x)^2 = (tan x)^2

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

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

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

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

cos x = 1/sqrt(25/144 + 1)

cos x = 1/sqrt(169/144)

cos x = sqrt(144/169)

**cos x = 12/13**