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Trigonometry Problem?
A wall extending east and west is 6 feet high. The sun has an altitude of 49 degrees 31’ and is 47 degrees 20’ east of south. Find the width of the shadow of the wall on level ground.
1 Answer
- V. SinghLv 69 years agoFavorite Answer
Consider wall as a pole of 6 feet high, If sun makes an angle of 49 degree 31' or 49 31/60 = 49.516°its shadow on the ground will be
|\
|...\
h|.....\
|.......\
|_____\
<--L---->
tan 49.516 = h / L
or L = (h = 6) / 1.1715 ; tan 49.516 = 1.1715
shadow (L) = 5.1215 feet
Now the sun is shifted towards east of south by 47° 20' or 47.3333° so shadow on the ground will be
shorter than this calculated as above
.... *|\
....*. |...\
...*...|.....\ 5.1215
.*.....|.......\
*......|_____\
<-----L------>
......<--w--->
So now shadow of wall will be w feet wide instead of L feet, which can be calculated as
cos 47.333 = w / 5.1215
or w = 0.6773*5.1215 = 3.47 feet ----- Answer
----------------------
It's bit difficult to explain without proper figure's in 3 D trigonometry, but I hope it would help you.
Vick