A overhead security camera needs to be placed so that both the far corner of a parking lot and an entry door are visible at the same time. The entry door is 23m from the camera, while the far corner of the parking lot is 19m from the camera. The far corner of the parking lot is 17m from the entry door. What angle of view for the camera, to the nearest degree is required?
Thanks so much guys!
Thanks so much guys!
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E----------------17 m -------------------P
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23m 19 m
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C
you have 3 sides of the triangle and need to solve for Angle C = Camera...
use the Law of Cosines...
17^2 = 23^2 + 19^2 - 2(23)(19)cosC
cosC = [17^2 - 23^2 - 19^2]/(-2(23)(19)] = 0.688
so C = cos^-1(0.688) = 47 degrees
Sorry I tried to draw the triangle but the editing was messed up when i hit submit
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23m 19 m
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- -
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C
you have 3 sides of the triangle and need to solve for Angle C = Camera...
use the Law of Cosines...
17^2 = 23^2 + 19^2 - 2(23)(19)cosC
cosC = [17^2 - 23^2 - 19^2]/(-2(23)(19)] = 0.688
so C = cos^-1(0.688) = 47 degrees
Sorry I tried to draw the triangle but the editing was messed up when i hit submit