A surveyor sights the top of a building with a handheld range finder. the top of the building is 148 feet 2 inches away. the ang
le of elevation is 56º. find the distance from the surveyor to the building in feet and inches. (sketch and label the situation first.)
1 answer:
see the attached figure to better understand the problem
we know that
in the right triangle ABC
cos 56°=AC/AB
where
AC is the adjacent side to angle 56 degrees------> the distance from the surveyor to the building
AB is the hypotenuse-----> 148 ft 2 in
56 degrees------> is the angle of elevation
so
cos 56°=AC/AB---------> solve for AC
AC=AB*cos 56°
AB=148 ft 2 in
convert 2 in to ft
1 ft -----> 12 in
x ft------> 2 in
x=2/12-----> x=0.17 ft
AB=148 ft 2 in-----> 148 ft+0.17 ft------> AB=148.17 ft
AC=AB*cos 56°----> AC=148.17*cos 56°------> AC=82.86 ft
convert 0.86 ft to in
0.86 ft=0.86*12-----> 10.32 in
distance AB=82 ft 10 in
the answer is
the distance from the surveyor to the building is 82 ft 10 in
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