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Answer :
see the attached picture to better understand the problem
we know that
in the right triangle ABC
tan 20°=AB/AC-----> AC=AB/tan 20----> AC=200/tan 20---> AC=549.50 ft
in the right triangle ABD
tan 12°=AB/AD----> AD=AB/tan 12----> AD=200/tan 12----> AD=940.93 ft
the runway is=DC
DC=AD-AC----> DC=940.93-549.50----> DC=391.43 ft---> 391 ft
the answer is
the runaway is 391 ft long
we know that
in the right triangle ABC
tan 20°=AB/AC-----> AC=AB/tan 20----> AC=200/tan 20---> AC=549.50 ft
in the right triangle ABD
tan 12°=AB/AD----> AD=AB/tan 12----> AD=200/tan 12----> AD=940.93 ft
the runway is=DC
DC=AD-AC----> DC=940.93-549.50----> DC=391.43 ft---> 391 ft
the answer is
the runaway is 391 ft long
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