Let's put more details in the figure to better understand the problem:
Let's first recall the three main trigonometric functions:
[tex]\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}[/tex][tex]\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}[/tex][tex]\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}[/tex]For x, we will be using the Cosine Function:
[tex]\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}[/tex][tex]Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}[/tex][tex](16)Cosine(45^{\circ})\text{ = x}[/tex][tex](16)(\frac{1}{\sqrt[]{2}})\text{ = x}[/tex][tex]\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}[/tex][tex]\text{ 8}\sqrt[]{2}\text{ = x}[/tex]Therefore, x = 8√2.
For y, we will be using the Sine Function.
[tex]\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}[/tex][tex]\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}[/tex][tex]\text{ (16)Sine }(45^{\circ})\text{ = y}[/tex][tex]\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}[/tex][tex]\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}[/tex][tex]\text{ 8}\sqrt[]{2}\text{ = y}[/tex]Therefore, y = 8√2.
