Given:
Find-:
The value of x and y
Explanation-:
In triangle ABC
[tex]\begin{gathered} AB=\text{ Base} \\ \\ BC=\text{ Perpendicular} \\ \\ AC=\text{ Hypotenuse} \end{gathered}[/tex]
Use trignometric property:
[tex]\begin{gathered} \sin\theta=\frac{\text{ Perpendicular}}{\text{ Hypotenuse}} \\ \\ \cos\theta=\frac{\text{ Base}}{\text{ Hypotenuse}} \end{gathered}[/tex]
For the value of "y" is:
[tex]\begin{gathered} \sin\theta=\frac{\text{ Perpendicular}}{\text{ Hypotenuse}} \\ \\ \sin60=\frac{BC}{AC} \\ \\ \sin60=\frac{5}{y} \\ \\ y=\frac{5}{\sin60} \end{gathered}[/tex]
Value of "Y" is:
[tex]\begin{gathered} y=\frac{5}{\sin60} \\ \\ y=\frac{5}{\frac{\sqrt{3}}{2}} \\ \\ y=\frac{10}{\sqrt{3}} \\ \\ y=\frac{10}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ y=\frac{10\sqrt{3}}{3} \end{gathered}[/tex]
The value of "x" is:
[tex]\begin{gathered} \cos60=\frac{\text{ Base}}{\text{ Hypotenuse}} \\ \\ \cos60=\frac{x}{y} \\ \\ \frac{1}{2}=\frac{x}{\frac{10\sqrt{3}}{3}} \\ \\ \frac{1}{2}=\frac{3x}{10\sqrt{3}} \\ \\ \frac{10\sqrt{3}}{2\times3}=x \\ \\ x=\frac{5\sqrt{3}}{3} \\ \\ \end{gathered}[/tex]
