The angle of elevation of the sun is 68° when a tree casts a shadow 14.3 m long. How tall is the tree, to the nearest tenth of a
meter?
1 answer:
See the attached picture to better understand the problem
we know that
in the right triangle ABC
tan 68°=opposite side angle 68/adjacent side angle 68
opposite side angle 68=AB----- height of the tree
adjacent side angle 68=BC-----> 14.3 m
so
tan 68=AB/BC-----> solve for AB
AB=BC*tan 68-----> AB=14.3*tan 68------> AB=35.39 m------> AB=35.4 m
the answer is
the height of the tree is 35.4 m
we know that
in the right triangle ABC
tan 68°=opposite side angle 68/adjacent side angle 68
opposite side angle 68=AB----- height of the tree
adjacent side angle 68=BC-----> 14.3 m
so
tan 68=AB/BC-----> solve for AB
AB=BC*tan 68-----> AB=14.3*tan 68------> AB=35.39 m------> AB=35.4 m
the answer is
the height of the tree is 35.4 m
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