Eric is standing 6 ft away from the base of the Empire State Building, which is 1453 ft tall. What is the angle of elevation when he looks a

Eric is standing 6 ft away from the base of the Empire State Building, which is 1453 ft tall. What is the angle of elevation when he looks at the top of the building? Round to the nearest hundredth.

$$\tan{ ( \mathrm{elevation\ensurement{^{\circ}} } ) } = \dfrac{ 1453 ~ft. }{ 6 ~ft.} = 242.1\overline{6} \\\\ \mathrm{elevation\ensurement{^{\circ}}} = \arctan{ ( 242.1\overline{6} ) }\\\\ \mathrm{elevation\ensurement{^{\circ}}} = 89.76 \ensurement{^{\circ}}$$