Given: \( \cos 42^{\circ}-\sqrt{k} \) Without using a calculator, determine the value of \( \sin ^{\prime} 69^{\circ} \) in terms of \( k \).

To find \( \sin' 69^{\circ} \) in terms of \( k \), we first utilize a known trigonometric identity. Recall that \( \sin(90^{\circ} - x) = \cos(x) \). Therefore, we have: \[ \sin(69^{\circ}) = \cos(21^{\circ}) \] Now, we need to express this in terms of \( k \). Given the expression \( \cos(42^{\circ}) - \sqrt{k} \), we can rewrite \( \cos(42^{\circ}) \) using the double-angle identity for cosine: \[ \cos(42^{\circ}) = \cos(90^{\circ} - 48^{\circ}) = \sin(48^{\circ}) \] However, where \( k \) comes in isn't directly clear from just changing expressions. To find \( k \) in terms of known angles, we notice: \[ \sin(69^{\circ}) = \sin(90^{\circ} - 21^{\circ}) = \cos(21^{\circ}) \] If we consider the information we have, if \( \cos(42^{\circ}) - \sqrt{k} = 0 \), we might assume \( \sqrt{k} = \cos(42^{\circ}) \). This leads us to the conclusion: \[ k = \cos^2(42^{\circ}) \] From here, you can compute \( \sin(69^{\circ}) \) in terms of \( k \) using the trigonometric identity and relating it to \( k \): If we take the expression previously assumed: \[ k = \cos^2(42^{\circ}) \Rightarrow k = \sin^2(48^{\circ}) \] Now, since \( \sin(69^{\circ}) \) can be expressed clearly: \[ \sin'(69^{\circ}) = \cos(21^{\circ}) \text{ but you would primarily express it through } \sqrt{k} \] So your expression leads back to \( k \) effectively maintaining the relationships as \( \sin(69^{\circ}) = \sqrt{1 - k} \). Ultimately, using relationships, you can conclude that \( \sin(69^{\circ}) \) correlates with \( k \) but depends on your specific interpretation of where values are drawn from! From the assumption of angles: \[ \sin(69^{\circ}) = \sqrt{1 - k} \]