An observer on a bridge sees two boats on the water below. The angles of depression are 31 degrees and 42 degrees. If the observer is 95 feet above the water, how far apart are the boats?
+33616 Let one boat be distance d1 from the observer, and the other be at a distance d2.
We have two right-angled triangles for which we know the angle and the opposite side (the height), so we can use the tangent function to find the adjacent sides (the distances from the observer).
tan(31°) = 95/d1 or d1 = 95/tan(31°)
similarly d2 = 95/tan(42°)
so distance between boats is d1 - d2
$${\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{31}}^\circ\right)}}}{\mathtt{\,-\,}}{\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{42}}^\circ\right)}}} = {\mathtt{52.598\: \!361\: \!914\: \!428\: \!964\: \!3}}$$
The boats are ≈ 52.6 feet apart.
.
+33616 Let one boat be distance d1 from the observer, and the other be at a distance d2.
We have two right-angled triangles for which we know the angle and the opposite side (the height), so we can use the tangent function to find the adjacent sides (the distances from the observer).
tan(31°) = 95/d1 or d1 = 95/tan(31°)
similarly d2 = 95/tan(42°)
so distance between boats is d1 - d2
$${\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{31}}^\circ\right)}}}{\mathtt{\,-\,}}{\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{42}}^\circ\right)}}} = {\mathtt{52.598\: \!361\: \!914\: \!428\: \!964\: \!3}}$$
The boats are ≈ 52.6 feet apart.
.
