Question
a pole leans away from the sun at an angle of 7 degrees to the vertical. when the elivation of the sun is 48 degrees the pole cast a shadow of 41ft long on level ground. how long is the pole?
Answers
a is vertically under pole, base to tip
41 - a is shadow tip from under pole tip
height of pole = y cos 7 = .993 y
tan 48 = .993 y/(41-a)
tan 7 = a/y so a = y tan 7
then
tan 48 = .993 y/(41 -y tan 7) solve for y
41 - a is shadow tip from under pole tip
height of pole = y cos 7 = .993 y
tan 48 = .993 y/(41-a)
tan 7 = a/y so a = y tan 7
then
tan 48 = .993 y/(41 -y tan 7) solve for y
Draw a big diagram
You end up with a triangle with:
the horizontal side = 47 ft
the side representing the leaning pole ... call it a
and the angle opposite the pole side = 51°
The pole makes an angle of 7° with the vertical ... so the angle between to pole side and the 47ft side = 90 - 7 = 83°
So that make the angle opposite the 47 ft side = 180 - (51 + 83) = 46°
Now you can use the sine rule to find the length of the pole:
a/sin51° = 47/sin46
a = 47 * sin51 / sin46
a = 50.8 ft
so the length of the pole is 50.8 ft
You end up with a triangle with:
the horizontal side = 47 ft
the side representing the leaning pole ... call it a
and the angle opposite the pole side = 51°
The pole makes an angle of 7° with the vertical ... so the angle between to pole side and the 47ft side = 90 - 7 = 83°
So that make the angle opposite the 47 ft side = 180 - (51 + 83) = 46°
Now you can use the sine rule to find the length of the pole:
a/sin51° = 47/sin46
a = 47 * sin51 / sin46
a = 50.8 ft
so the length of the pole is 50.8 ft
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