First get the adjacent angle of 17˚
180˚ - 17˚ = 163˚
Then get the third angle of the triangle formed by the vertex of the 15˚-angle, vertex of 163˚ angle and the top of tree.
180˚ - 15˚ - 163˚ = 2˚
Then get the length from the vertex of the 15˚-angle to the top of the tree using sine law.
31.4 / sin 2˚ = x / sin 163˚
x = 31.4*sin 163˚ / sin 2˚
x = 263.1 ft
Finally get the height of the tree (h) using sine (or any trigonometric identities taking into account the angle to be used.
sine 15˚ = h / 263.1
h = 263.1 * sin 15˚
h = 68.1 ft
So the height of the tree is approximately 68.1 ft.
2006-12-19 16:22:02
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answer #1
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answered by dkrudge 2
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68.1 is your tree height.
You MIGHT want to look at the right triangle(s) taking the 2 acute angles at the top of the tree (73 deg, and 75 deg)
tan 73 = x / h where x is the distance from tree base to vertex of 17 deg angle, and h = tree height
tan 75 = (x+31.4) / h
Eliminate x from the 2 equations; solve for h.
(FWIW x = 222.7 feet; so x + 31.4 = 254.1)
2006-12-22 09:10:50
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answer #2
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answered by answerING 6
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Let the height of the tree be h and the distance from the base to the farther most point (where angle is 15) be x ...
Tan 15 = h / x .... x = h / Tan 15
Tan 17 = h / (x-31.4) ........x-31.4 = h /Tan 17 ............. x = h/Tan 17 + 31.4
So h/Tan 15 = (h/Tan 17) + 31.4
h/Tan 15 - h/Tan 17 = 31.4
h = 31.4 (Tan 15)(Tan 17) / (Tan 17 - Tan 15)
Substitute Tan 15 and Tan 17 values to get h, the height of the tree.
2006-12-19 16:42:16
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answer #3
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answered by Srinivas c 2
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If performed in a vacuum that is obviously actual. The bullet's vertical acceleration in each instances is brought about thoroughly via gravity because no drive as opposed to gravity has acted at the bullet within the vertical course. The horizontal movement brought about via the firing of the bullet is not going to difference the vertical acceleration because of gravity, and as a consequence the dropped bullet and fired bullet will hit the bottom whilst. Note that virtually, the aerodynamic results at the bullet may not make this particularly actual. The forces of elevate and drag at the fired bullet vs. the dropped bullet will range, that means one will hit the bottom earlier than the opposite. Without jogging the numbers I'd bet the dropped bullet will hit first, because it'll certainly be orientated up and down, exposing much less floor subject to tug, however on the speed the fired bullet is touring a extra intensive evaluation could be critical to be definite.
2016-09-03 14:54:06
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answer #4
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answered by Anonymous
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Let
h= height of tree
x = horizontal distance to tree from 17° angle
x + 31.4 = horizontal distance to tree from 15° angle
tan = opposite side / adjacent side
tan 17° = h / x
tan 15° = h / (x + 31.4)
x = h / tan 17°
x + 31.4 = h / tan 15°
x = h / tan 15° - 31.4
h / tan 17° = h / tan 15° - 31.4
h / tan 17° - h / tan 15° = -31.4
h / tan 15° - h / tan 17° = +31.4
h * tan 17° - h * tan 15° = 31.4(tan 15°)(tan 17°)
h(tan 17° - tan 15°) = 31.4(tan 15°)(tan 17°)
h = 31.4(tan 15°)(tan 17°)/(tan 17° - tan 15°)
h = 68.1 feet
2006-12-19 18:03:36
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answer #5
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answered by Northstar 7
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let a and b be the length from the leg of tree to the point of each angle
h is the height of the tree
a + 31.4 = b
we have : tan 17 / tan15 = b / a <=> a * tan17 = b*tan15 = (a + 31.4)*tan15
=> a =31.4 / (tan17 - tan15) = 831 ft
h = a * tan17 = 254 ft
2006-12-19 17:54:17
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answer #6
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answered by James Chan 4
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let the heightbe h andthe horizontal dist be x
h=xtan15*
h=(x+31.4)tan 17*
solve for x
substitute toget h
2006-12-19 16:07:14
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answer #7
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answered by raj 7
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831'-1 1/8" or 68'-1"
2006-12-19 16:24:24
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answer #8
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answered by Tim F 1
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