mathematics-geometry

[匿名]

2011-11-01 3:16 am

Triangle ABC and BED are both equilateral triangles. Given 2 other angles, find Angle CDE. (Sin, cos, tan are forbidden, only other other geometry formulas which are up to grade 9-form 3)

圖片參考：http://imgcld.yimg.com/8/n/HA07450594/o/701110310069913873394730.jpg

In the graph below, mark that the 4 sides are not equal, you already know all interior angles except Angles DAO and ADO (O is the centre crossed spot by two lines), find them 'individually' in term of other known angles (Sin, cos, tan are forbidden, only other other geometry formulas which are up to grade 9-form 3)

圖片參考：http://imgcld.yimg.com/8/n/HA07450594/o/701110310069913873394741.jpg

更新1:

I mean in question 2 ADO and OAD dont related in both equations, meaning to show them individually, not ADO in DAO and DAO in ADO. By the way ,thanks to the 1. question

回答 (1)

用戶2053

2011-11-01 4:35 am

✔ 最佳答案

1)

BE = BD (equilateral triangle sides)
BC = BA (equilateral triangle sides)
ㄥEBC = ㄥEBD - ㄥCBD = 60° - ㄥCBD = ㄥCBA - ㄥCBD = ㄥDBA

∴ △BEC ≡ △BDA (S.A.S.)

So ㄥBEC = ㄥBDA = 95°
Therefore ㄥDEC = ㄥBEC - ㄥBED = 95° - 60° = 35°
3)

圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701110310069913873394740.jpg

Extend BD and BC to E and F respectively such that EF // DC ,

then all known angles remain unchange except ㄥDAO and ㄥADO .

So we can only conclude that

ㄥDAO + ㄥADO + ㄥCAB + ㄥABC + ㄥBCD + ㄥCDB = 360°

i.e.

ㄥDAO = 360° - ㄥADO - ㄥCAB - ㄥABC - ㄥBCD - ㄥCDB

ㄥADO = 360° - ㄥDAO - ㄥCAB - ㄥABC - ㄥBCD - ㄥCDB

2011-11-02 17:28:45 補充：
Corrections :

In Q2)

We have to use sin or cos formula to find them since 4 lines have no special relation.

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