Question 1142774: A pole casts a shadow of 15 m when the angle of elevation of the sun is 61 degrees.
If the pole has leaned 15 degrees from the vertical directly toward the sun, what is the length of the pole?
Assuming that the triangle formed is a plane figure. What is the area of the figure using the heron's formula? Thanks.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! This is an application of the Law of Sines
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let b be the length of the pole
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the triangle formed has the following properties(label the angles A, B, C beginning with the top angle and proceed clockwise)
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Angle A is 14 degrees, Angle B is 61 degrees and angle C is 105 degrees(90+15)
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Note a is the length of the shadow, 15 m, then we have
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b/sin(61) = 15/sin(14)
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b = (15 * sin(61))/sin(14) = 54.23 m
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length of pole is 54.23 m
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c/sin(105) = 15/sin(14)
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c = (15 * sin(105))/sin(14) = 59.89 m
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Heron's formula
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s = (1/2) * (a + b + c) = (1/2) * (15 + 54.23 + 59.89) = 64.56
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Area of triangle = square root(s * (s - a) * (s - b) * (s - c)) = square root (64.56 * (64.56 - 15) * (64.56 - 54.23) * (64.56 - 59.89)) = 392.88 m^2
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