## Physics (10th Edition)

We look for the distance between the top of the building and the reference point $tan(35^{\circ}) = \frac{x}{85}$ $x \approx 59.5176\; m$ We look for the distance between the top of the antenna and reference point $tan(38^{\circ}) = \frac{x+h}{85}$ $x+h \approx 66.4093\; find difference 66.4093-59.5176$ = 6.89 m
$tan(35^{\circ}) = \frac{x}{85}$ $x \approx 59.5176\; m$ $tan(38^{\circ}) = \frac{x+h}{85}$ $x+h \approx 66.4093\; m$ $(59.5176) + h = 66.4093$ $h = 66.4093 - 59.5176 \approx 6.89 m$