題目:
In the figure, Sammi and Andy are 170m apart. There is a lamp-post between them. Sammi, Andy and the lamp-post form a straight line. Find the height of the lamp-post.
Figure:
~~~~~~~~~~~~~~top of the lamp-post
~~~~~~~~~~／~~~~~~~~~~~~~~~~~＼~~~~~~
~~~~~~~~／~~~~~~~~~~~~~~~~~~~~~＼~~~
~~~~~~／~~~~~~~~~~~~~~~~~~~~~~~~~＼~~~~~
Andy∠35°~~~~~~~~Lamp-post~~~~~~~~40°＼Sammi
老師話要用sin cos tan
唔識計
數學難題 唔識做
2009-06-03 6:52 am
回答 (4)
2009-06-03 9:42 am
✔ 最佳答案
Let (X)m be the length between Andy and Lamp-post,(170-X)m be the length between Sammi and Lamp-post,
(Y)m be the height of the lamp-post.
Because Y/X = tan35
Y = tan35 x X
and Y/(170-X) = tan40
Y = tan40 x (170-X)
So tan35 x X = tan40 x (170-X)
0.700 x X = 0.839 x (170-X)
0.700X = 0.839 x 170 - 0.839 x X
0.700X = 142.63 - 0.839X
0.700X + 0.839X = 142.63
1.539X = 142.63
X = 92.677
Height of Lamp-post:
Y/X = tan35
Y/92.677 = tan35
Y/92.677 = 0.700
Y = 0.700 x 92.677
Y = 64.87
So, the height of the lamp-post is 64.87m.
2009-06-04 10:33 pm
Mathod 1
Let Point A is Andy
Let Point L is Lamp post
Let Point S is Sammi
Let Point T is the top of the lamp post
Then ATS is a triangle with angle A = 35° and angle S is 40°
Also TL bisect the triangle into two right angle triangles at T
Hence we have two equations
tan 35 = TL / AL ...............1
tan 40 = TL / (170-AL)......2
from 1 AL = TL / tan 35
from 2 AL = 170 - TL / tan 40
therefore TL / tan 35 = 170 - TL / tan 40
TL tan 40 = 170 x tan35 x tan 40 - TL tan 35
TL (tan 40 + tan 35) = 170 x tan 35 x tan 40
TL = 170 x tan 35 x tan 40 / (tan 40 + tan 35)
Mathod 2
Let (X)m be the length between Andy and Lamp-post,
(170-X)m be the length between Sammi and Lamp-post,
(Y)m be the height of the lamp-post.
Because Y/X = tan35
Y = tan35 x X
and Y/(170-X) = tan40
Y = tan40 x (170-X)
Therefore tan35 x X = tan40 x (170-X)
0.700 x X = 0.839 x (170-X)
0.700X = 0.839 x 170 - 0.839 x X
0.700X = 142.63 - 0.839X
0.700X + 0.839X = 142.63
1.539X = 142.63
X = 92.677
Height of Lamp-post:
Y/X = tan35
Y/92.677 = tan35
Y/92.677 = 0.700
Y = 0.700 x 92.677
Y = 64.87
Therefore, the height of the lamp-post is 64.87m.
Let Point A is Andy
Let Point L is Lamp post
Let Point S is Sammi
Let Point T is the top of the lamp post
Then ATS is a triangle with angle A = 35° and angle S is 40°
Also TL bisect the triangle into two right angle triangles at T
Hence we have two equations
tan 35 = TL / AL ...............1
tan 40 = TL / (170-AL)......2
from 1 AL = TL / tan 35
from 2 AL = 170 - TL / tan 40
therefore TL / tan 35 = 170 - TL / tan 40
TL tan 40 = 170 x tan35 x tan 40 - TL tan 35
TL (tan 40 + tan 35) = 170 x tan 35 x tan 40
TL = 170 x tan 35 x tan 40 / (tan 40 + tan 35)
Mathod 2
Let (X)m be the length between Andy and Lamp-post,
(170-X)m be the length between Sammi and Lamp-post,
(Y)m be the height of the lamp-post.
Because Y/X = tan35
Y = tan35 x X
and Y/(170-X) = tan40
Y = tan40 x (170-X)
Therefore tan35 x X = tan40 x (170-X)
0.700 x X = 0.839 x (170-X)
0.700X = 0.839 x 170 - 0.839 x X
0.700X = 142.63 - 0.839X
0.700X + 0.839X = 142.63
1.539X = 142.63
X = 92.677
Height of Lamp-post:
Y/X = tan35
Y/92.677 = tan35
Y/92.677 = 0.700
Y = 0.700 x 92.677
Y = 64.87
Therefore, the height of the lamp-post is 64.87m.
參考: me
2009-06-03 7:24 am
題目係咪仲有些條件未寫?
2009-06-03 7:12 am
Let Point A is Andy
Let Point L is Lamp post
Let Point S is Sammi
Let Point T is the top of the lamp post
Then ATS is a triangle with angle A = 35° and angle S is 40°
Also TL bisect the triangle into two right angle triangles at T
Hence we have two equations
tan 35 = TL / AL ...............1
tan 40 = TL / (170-AL)......2
from 1 AL = TL / tan 35
from 2 AL = 170 - TL / tan 40
therefore TL / tan 35 = 170 - TL / tan 40
TL tan 40 = 170 x tan35 x tan 40 - TL tan 35
TL (tan 40 + tan 35) = 170 x tan 35 x tan 40
TL = 170 x tan 35 x tan 40 / (tan 40 + tan 35)
Let Point L is Lamp post
Let Point S is Sammi
Let Point T is the top of the lamp post
Then ATS is a triangle with angle A = 35° and angle S is 40°
Also TL bisect the triangle into two right angle triangles at T
Hence we have two equations
tan 35 = TL / AL ...............1
tan 40 = TL / (170-AL)......2
from 1 AL = TL / tan 35
from 2 AL = 170 - TL / tan 40
therefore TL / tan 35 = 170 - TL / tan 40
TL tan 40 = 170 x tan35 x tan 40 - TL tan 35
TL (tan 40 + tan 35) = 170 x tan 35 x tan 40
TL = 170 x tan 35 x tan 40 / (tan 40 + tan 35)
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