Answer:
16.06 ft
Step-by-step explanation:
The figure is attached below.
In triangle ACB:
[tex]tan(41)=\frac{x}{y} \\x=ytan(41)[/tex]
In triangle ADB:
[tex]tan(25)=\frac{x}{y+10} \\(y+10)tan(41)=x[/tex]
Therefore equating both equations gives:
[tex]ytan(41) = (y+10)tan(25)\\ytan(41) = ytan(25)+10tan(25)\\ytan(41)-ytan(25)=10tan(25)\\y(tan(41)-tan(25))=10tan(25)\\y=\frac{10tan(25)}{(tan(41)-tan(25)} =11.5715ft[/tex]
Therefore x = 11.5715*tan(41) = 10.06 ft
The distance of the jot air balloon to ground = 10.06 + 6 = 16.06 ft
