In any right triangle with acute angles x and y, then
The sum of x and y is 90 degrees
[tex]\begin{gathered} \sin x=\cos y \\ \cos x=\sin y \\ x+y=90^{\circ} \end{gathered}[/tex]
Then for part (1)
Since triangle XYZ is a right angle at Z
Then
[tex]X+Y=90^{\circ}[/tex]
Then X and Y are complementary angles
Part (2)
sin X = opposite/hypotenuse
[tex]\sin X=\frac{x}{z}[/tex]
sin Y = opposite/hypotenuse
[tex]\sin Y=\frac{y}{z}[/tex]
cos X = adjacent/hypotenuse
[tex]\cos X=\frac{y}{z}[/tex]
cos Y = adjacent/hypotenuse
[tex]\cos Y=\frac{x}{z}[/tex]
Part (3)
[tex]\begin{gathered} \sin X=\cos Y \\ \cos X=\sin Y \end{gathered}[/tex]
Part (4)
Since sin = cos, then
The sum of the 2 angles must be 90
One of them is 23 degrees, then the other must be
[tex]90-23=67[/tex]
The answer is
[tex]\cos (23)=\sin (67)[/tex]
