Answer:
The answer is "[tex]\bold{17456.6 \ ft}[/tex]"
Step-by-step explanation:
Please find the complete question in the attached file.
[tex]\sin \theta = \frac{P}{h}\\\\\to BC =x\\\\\to BD =y\\\\[/tex]
The height of tower (AB) = 300 ft
Distance = CD
[tex]\Delta ABC \\\\\sin \theta = \frac{AB}{AC}\\\\\sin (2.5) = \frac{300}{AC}\\\\Ac= \frac{300}{\sin (2.5)}\\\\AC= 6877.68 \\\\\Delta ABD \\\\\sin \theta = \frac{AB}{AD}\\\\\sin (1.3) = \frac{300}{AD}\\\\AD= \frac{300}{\sin (1.3)}\\\\AC= 1322.33[/tex]
calculating CD
:
[tex]CD^2 = AC^2 +AD^2 -2\cdot AC \cdot AD \cdot \cos(117^{\circ}) \\\\CD = \sqrt{(6877.68)^2 +(13223.23)^2 -2\cdot (6877.68) \cdot (13223.23) \cdot \cos(117^{\circ})}\\\\CD =\sqrt{47302482.1824 + 174853811.6329 - 82576463.207}\\\\CD= \sqrt{304732757.02}\\\\CD= 17456.6 \ ft[/tex]
