Answer:
x = [tex]\frac{16}{\sqrt{2} }[/tex] and y = [tex]\frac{16}{\sqrt{2} }[/tex]
Step-by-step explanation:
Applying the trigonometric functions to the given triangle;
i. To determine the value of x;
Sin θ = [tex]\frac{opposite}{hypotenuse}[/tex]
Sin 45 = [tex]\frac{x}{16}[/tex]
x = 16 * Sin 45
= 16 * [tex]\frac{1}{\sqrt{2} }[/tex]
x = [tex]\frac{16}{\sqrt{2} }[/tex]
ii. To determine the value of y;
Cos θ = [tex]\frac{adjacent}{hypotenuse}[/tex]
Cos 45 = [tex]\frac{y}{16}[/tex]
y = 16 * Cos 45
= 16 * [tex]\frac{1}{\sqrt{2} }[/tex]
y = [tex]\frac{16}{\sqrt{2} }[/tex]
Thus, x = [tex]\frac{16}{\sqrt{2} }[/tex] and y = [tex]\frac{16}{\sqrt{2} }[/tex]
