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20 13 Tianjin mathematics exam round questions plus answers.
Solution: (1) As shown in Figure ①, connect OC,

Lines l and o are tangent to point c,

∴OC⊥l,

∵AD⊥l,

∴OC∥AD,

∴∠OCA=∠DAC,

OA = OC,

∴∠BAC=∠OCA,

∴∠bac=∠dac=30;

(2) As shown in Figure ②, connect BF,

∵AB is the diameter⊙ O,

∴∠AFB=90,

∴∠BAF=90 -∠B,

∴∠aef=∠ade+∠dae=90+ 18 = 108

In ⊙O, the quadrilateral ABFE is the inscribed quadrilateral of a circle,

∴∠AEF+∠B= 180,

∴∠B= 180 - 108 =72,

∴∠BAF=90 -∠B=90 -72 = 18。