多邊形(數學)

2)
In △ABC
∵AB=AC (given)
∴∠ABC=∠ACB (base.∠s, isos.△)
∠ACB=(180°-50°)/2=65°
In△ACD
50°+∠ADC+∠ACD=180° (∠ sum of △)
50°+(180°-90°)+∠ACD=180°
∠ACD=40°
∴∠BCD=∠ACB-∠ACD
∠BCD=65°-40°=25°

3)
Each of the interior ∠ in the polygon=[(n-2)x180°]/6 , where n is the number of size
∠AFE=120°
In △ABF
∵AF=AB (given)
∴∠AFB=∠ABF (base.∠s, isos.△)
∠AFB=(180°-120°)/2=30°
In △DEF
∵DE=EF (given)
∴∠EFD=∠EDF (base.∠s, isos.△)
∠EFD=(180°-120°)/2=30°
∴x+∠EFD+∠AFB=∠AFE
x+30°+30°=120°
x=60°

2007-08-15 21:35:11 補充：
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( 2 ) ∠ DAC + ∠ ACD = 90* ( ext. ∠ of triangle )

50* + ∠ ACD = 90*

∠ ACD = 40*

∠ ABC = ∠ ACB ( base ∠s, isos. triangle )

2 ∠ ACB + 50* = 180* ( ∠ sum of triangle )

∠ ACB = 65*

∠ BCD = 65* - 40* = 25*

( 3 ) For a regular polygon, all sides are equal.

The size of each interior angle :

( 6 - 2 ) x 180* / 6

= 120*

By SAS, triangles FED and ABF are concurrent.

∠ AFB = ∠ ABF ( base ∠s, isos. triangle )

120* + 2 ∠ AFB = 180* ( ∠ sum of triangle )

∠ AFB = 30*

Then ∠ EFD = 30* ( corr. ∠s, concurrent triangles )

x + 30* + 30* = 120*

x = 60*