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(1) ∠x を求めなさい.
66° 41° x 28° |
(2) ∠x を求めなさい.
35° x 42° 52° |
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(3) ∠x を求めなさい.
x A 29° B 25° C 134° D E F |
(4) ∠x を求めなさい.
65° 35° 65° x |
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(1) ∠x を求めなさい.
66° 41° x 28° 対頂角 = 180−(∠x+41) = 180−(66+28) ∠x = 53 |
(2) ∠x を求めなさい.
35° x 42° 52° 対頂角 = 180−(∠x+42) = 180−(35+52) ∠x = 45 |
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(3) ∠x を求めなさい.
x A 29° B 25° C 134° D E F よって,∠DFB = ∠BDC − ∠B = 134 − 29 = 105 ∠DFBは△AFCの頂点Fにおける外角だから∠DFB = ∠A+∠C ∠DFB = ∠x + ∠C = 105 よって,∠x = 80 |
(4) ∠x を求めなさい.
65° 35° 65° x 対頂角 = 180−(∠x+65) = 180−(35+65) ∠x = 35 |