the pilot of a helicopter sights a fire at an angle of depression of 11 degrees. the helicopters altitude is 1700 feet.to the nearest foot, what is the horizontal distance from the helicopter to the fire?
+33666 tan(angle of depression) = height/horizontal distance
so horizontal distance = height/tan(angle of depression)
$${\frac{{\mathtt{1\,700}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{11}}^\circ\right)}}} = {\mathtt{8\,745.741\: \!827\: \!136\: \!861\: \!007\: \!1}}$$
horizontal distance ≈ 8746 ft
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+33666 tan(angle of depression) = height/horizontal distance
so horizontal distance = height/tan(angle of depression)
$${\frac{{\mathtt{1\,700}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{11}}^\circ\right)}}} = {\mathtt{8\,745.741\: \!827\: \!136\: \!861\: \!007\: \!1}}$$
horizontal distance ≈ 8746 ft
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