1996 AMC 8 第 24 题

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24.

ABCABC 的度数是 5050^\circAD\overline{AD} 平分角 BACBACDC\overline{DC} 平分角 BCABCA。角 ADCADC 的度数是

The measure of angle ABCABC is 5050^\circ. AD\overline{AD} bisects angle BACBAC, and DC\overline{DC} bisects angle BCABCA. The measure of angle ADCADC is

9090^\circ

100100^\circ

115115^\circ

122.5122.5^\circ

125125^\circ

答案:C
知识点:导角角平分线
难度评级:1150
解答:

在三角形 ABCABC 中,BAC+BCA=18050\angle BAC + \angle BCA = 180^\circ - 50^\circ =130= 130^\circ

角平分线给出 DAC+DCA=1302=65\angle DAC + \angle DCA = \tfrac{130^\circ}{2} = 65^\circ。在三角形 ADCADC 中,ADC=18065=115\angle ADC = 180^\circ - 65^\circ = 115^\circ

所以正确答案是 C

In triangle ABCABC, BAC+BCA=18050\angle BAC + \angle BCA = 180^\circ - 50^\circ =130= 130^\circ.

The bisectors give DAC+DCA=1302=65\angle DAC + \angle DCA = \tfrac{130^\circ}{2} = 65^\circ. In triangle ADCADC, ADC=18065=115\angle ADC = 180^\circ - 65^\circ = 115^\circ.

Thus, the correct answer is C .

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