Asked by AEF
A straight road makes an angle, A of 12 degrees. When the angle of elevation, B, of the sun is 55 degrees, a vertical pole beside the road casts a shadow 7 feet long parallel to the road. Approximate the length of the pole. Round to tow decimal places.
Answers
Answered by
Steve
draw a diagram.
P = end of shadow on road
T = top of pole
Q = foot of pole
Z = intersection of horizontal line from Q and pole extended vertically.
Let d = PT: the distance from the tip of the shadow to the top of the pole
Let x = PZ
Let y = ZQ
Let h = QT, the height of the pole
We know:
∠ZPT = 55°
∠ZPQ = 12°
so, ∠QPT = 43°
PQ = 7
x = 7cos12° = 6.847
y = 7sin12° = 1.455
d = x/cos55° = 7cos12°/cos55° = 11.937
now, we can do this two ways
(1) Pythagorean Theorem
x^2 + (y+h)^2 = d^2
h = 8.323
(2) Law of Cosines
h^2 = d^2 + 7^2 - 2*7*d*cos43°
h = 8.323
P = end of shadow on road
T = top of pole
Q = foot of pole
Z = intersection of horizontal line from Q and pole extended vertically.
Let d = PT: the distance from the tip of the shadow to the top of the pole
Let x = PZ
Let y = ZQ
Let h = QT, the height of the pole
We know:
∠ZPT = 55°
∠ZPQ = 12°
so, ∠QPT = 43°
PQ = 7
x = 7cos12° = 6.847
y = 7sin12° = 1.455
d = x/cos55° = 7cos12°/cos55° = 11.937
now, we can do this two ways
(1) Pythagorean Theorem
x^2 + (y+h)^2 = d^2
h = 8.323
(2) Law of Cosines
h^2 = d^2 + 7^2 - 2*7*d*cos43°
h = 8.323
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