Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 281: 50b

The opposite side is $x$ The adjacent side is $\sqrt{1-x^2}$ The hypotenuse is 1

$u = arcsin~x$ By the definition of arcsin: $x = sin~u$ $sin~u = x = \frac{x}{1} = \frac{opp}{hyp}$ We can find an expression for the adjacent side: $adj = \sqrt{1^2-x^2} = \sqrt{1-x^2}$ The opposite side is $x$ The adjacent side is $\sqrt{1-x^2}$ The hypotenuse is 1