# If tan X = 16/18, can the value of sin X be found.

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### 2 Answers

It is given that `tan X = 16/18` .

`tan X = (sin X)/(cos X)`

Use `sin^2 X + cos^2 X = 1`

=> `cos^2 X = 1 - sin^2X`

=> `cos X = sqrt(1 - sin^2 X)`

`tan X = (sin X)/(cos X)`

=> `16/18 = (sin X)/(sqrt(1 - sin^2 X)) `

=> `16*sqrt(1 - sin^2X) = 18*sin X`

=> `256*(1 - sin^2X) = 324*sin^2X`

=> `256 - 256*sin^2X = 324*sin^2X`

=> `sin^2X = 256/580`

=> `sin X = 16/sqrt 580` and `sin X = -16/sqrt 580`

**There can be two values of sin X, `16/sqrt 580` and **`-16/sqrt 580`

if tanX = 16/18, can value of X be found?

Tan of an angle is Perpendicual divided by base for the angle of a right-angled triangle at base.

Let "k" be multiplier so that

In this case, the base = 18k and perpendicular = 16k

Fo above,

The hypotenuse of the triangle = sqrt((16k)^2+(18k)^2)

= sqrt(256+324).k = sqrt(580)k

sinX = perpendicular/hypotenuse

sinX = 16/sqrt(580)k = 8/sqrt(145) = 0.66436 approx.

**The value of sinX = 8/sqrt(145) **or** 0.66436 **when tanX = 16/18