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Old 07-26-2007   #1
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Default trigonometry help please... i have a test tomorrow... help?

a 6m treee is leaning at an angle of 5* to the vertcal.to prevent the tree from leaning any farther, a support pole neeeds to be placed 2m from the top of the tree, at an angle of 60* with the ground.how long does the pole need to be, to the nearest meter?please show all calculations and the final answer.
Old 07-26-2007   #2
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Draw a triangle ABC, where A= point where pole meets ground, B=point where pole meets tree, C=point where tree meets ground.Angle ACB = 85 deg, since tree is 5 degrees from verticalAngle BAC = 60 degrees, givenAngle ABC = 180 - 85 - 60 = 35 deg, since interior angles of a triangle add up to 180 degSide BC = 6 - 2 = 4m, since pole meets tree 2m from its top.There are several ways to proceed from here; one way is to use the law of sines, wherec / sin(C) = b / sin(B) = a / sin(A)So, c / sin(85) = 4 / sin(60)c = 4 (2/ sqrt(3)) (sin(85) = (8 /sqrt(3)) sin(85)or c = 4.6(Check for math errors above, but the logic should be OK)
Old 07-26-2007   #3
Frank K
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Answer = 4.60 meters or 5 meters to nearest whole meter.Take a triangle ABC with side AB = 4 m,
Old 07-26-2007   #4
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2 m from the top of the tree is an altitude of about 4 m (more precisely, 3.985 m, which is 4 m multiplied by the cosine of 5 degrees). This altitude is equal to the length of the pole multiplied by the sine of 60 degrees.In other words, the length of the pole is4 m) cos(5) / sin(60) = 4.6012... mTo the nearest meter, that's 5 m.

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