[tex]\begin{gathered} \text{To find A} \\ A+119+27=180 \\ A=180-119-27 \\ A=34 \\ U\sin g\text{ sine theorem} \\ \frac{AB}{\sin(27)}=\frac{BC}{\sin(34)} \\ \text{Solving AB} \\ AB=\frac{BC\sin(27)}{\sin(34)} \\ BC=250 \\ AB=\frac{250\sin(27)}{\sin(34)} \\ AB=202.966\approx203 \\ \text{The distance AB is 203} \\ \\ \text{Question 2} \\ To\text{ find the }height\text{ of the h}ill \\ height\text{ = OB} \\ \tan (10)=\frac{OB}{600} \\ \text{Solving OB} \\ OB=600\tan (10) \\ OB=105.796\text{ fe}et\approx106\text{ fe}et \\ \text{The height of the hill is 106 ft} \\ To\text{ find the height of the antenna} \\ \text{Antenna}=OT-OB \\ \tan (25)=\frac{OT}{600} \\ \text{Solving OT} \\ OT=600\tan (25) \\ OT=279.78\text{feet}\approx279.8feet \\ \text{Hence} \\ \text{Antenna}=279.8feet-106\text{ fe}et \\ \text{Antenna}=137.8\text{ fe}et \\ \text{The height of the antenna is 137.8fe}et \end{gathered}[/tex]
