The given diagram is
Recall the sine formula for the given triangle is
[tex]\frac{150}{\sin A}=\frac{y}{\sin B}=\frac{x}{\sin C}[/tex]
Substitute the angles as follws.
[tex]\frac{150}{\sin A}=\frac{y}{\sin62^o}=\frac{x}{\sin38^o}[/tex]
By using the triangle sum property, we get
[tex]\angle A+\angle B+\angle C=180^o[/tex]
Substitute known values, we get
[tex]\angle A+62^o+38^o=180^o[/tex]
[tex]\angle A=180^o-100^o[/tex][tex]\angle A=80^o[/tex]
[tex]\frac{150}{\sin 80^o}=\frac{y}{\sin62^o}=\frac{x}{\sin38^o}[/tex]
[tex]\text{Consider }\frac{150}{\sin 80^o}=\frac{y}{\sin62^o}\text{.}[/tex][tex]\text{Use the values }\sin 80^o=0.98\text{ and }\sin 62^o=0.88.[/tex]
[tex]\text{ }\frac{150}{0.98}=\frac{y}{0.88}[/tex]
[tex]y=\text{ }\frac{150\times0.88}{0.98}[/tex][tex]y=134.69\text{ f}eet[/tex]
[tex]\text{Consider }\frac{150}{\sin 80^o}=\frac{x}{\sin 38^o}[/tex]
[tex]\text{Use the values }\sin 80^o=0.98\text{ and }\sin 38^o=0.62.[/tex][tex]\frac{150}{0.98}=\frac{x}{0.62}[/tex]
[tex]x=\frac{150\times0.62}{0.98}[/tex][tex]x=94.90\text{ fe}et[/tex]
Hence the length of x and y is
[tex]x=94.9\text{ fe}et[/tex]
[tex]y=134.7\text{ f}eet[/tex]
