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(1) ∠x を求めなさい.
55° A 31° B x C 135° D E F |
(2) ∠x を求めなさい.
65° A 29° B 34° C x D E |
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(3) ∠x を求めなさい.
75° A x B 33° C 135° D E F |
(4) ∠x を求めなさい.
39° x 52° 52° |
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(1) ∠x を求めなさい.
55° A 31° B x C 135° D E F よって,∠DFB = ∠BDC − ∠B = 135 − 31 = 104 ∠DFBは△AFCの頂点Fにおける外角だから∠DFB = ∠A+∠C ∠DFB = ∠A + ∠x = 104 よって,∠x = 49 |
(2) ∠x を求めなさい.
65° A 29° B 34° C x D E ∠BEC = 65 + 29 = 94 ∠BDCは△DCEの頂点Dにおける外角だから∠BDC = ∠C+∠BEC ∠BDC = 34+94 = 128 |
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(3) ∠x を求めなさい.
75° A x B 33° C 135° D E F ∠CFB = 75 + 33 = 108 ∠BDCは△BDFの頂点Dにおける外角だから∠BDC = ∠B+∠CFB ∠B = ∠x = ∠BDC − ∠CFB = 27 |
(4) ∠x を求めなさい.
39° x 52° 52° 対頂角 = 180−(∠x+52) = 180−(39+52) ∠x = 39 |