Answer :
Answer:
Explanation:
<ADC = 30.5 Degree
<BDC = 21.5 Degree
AB = 133 ft
[tex]in \Delta BCD[/tex]
[tex]tan 21.5= \frac{BC}{CD}[/tex]
[tex]CD = \frac{BC}{tan21.5}[/tex]
in[tex] \Delta ACD[/tex]
[tex]tan 30.5 =\frac{AC}{CD}[/tex]
[tex]CD = \frac{AB+BC}{tan30.5}[/tex]
[tex] \frac{BC}{tan21.5} = \frac{AB}{tan30.5} +\frac{BC}{tan30.5}[/tex]
[tex]\frac{BC}{tan21.5} - \frac{BC}{tan30.5} = 225.78[/tex]
2.5BC - 1.69BC = 225.78
BC = 281.40 ft
