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(1) ∠x を求めなさい.
32° 42° x 51° |
(2) ∠x を求めなさい.
x 40° 40° 40° |
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(3) ∠x を求めなさい.
65° A 29° B x C 135° D E F |
(4) ∠x を求めなさい.
44° 49° x 49° |
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(1) ∠x を求めなさい.
32° 42° x 51° 対頂角 = 180−(∠x+42) = 180−(32+51) ∠x = 41 |
(2) ∠x を求めなさい.
x 40° 40° 40° 対頂角 = 180−(∠x+40) = 180−(40+40) ∠x = 40 |
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(3) ∠x を求めなさい.
65° A 29° B x C 135° D E F よって,∠DFB = ∠BDC − ∠B = 135 − 29 = 106 ∠DFBは△AFCの頂点Fにおける外角だから∠DFB = ∠A+∠C ∠DFB = ∠A + ∠x = 106 よって,∠x = 41 |
(4) ∠x を求めなさい.
44° 49° x 49° 対頂角 = 180−(∠x+49) = 180−(44+49) ∠x = 44 |