Question

如图AB是半圆O的直径，C为半圆上一点，角CAB的平分线AE交BC于点D，交半圆O于点E,若AB=10，tan角CAB=3/4

求线段CD的长

Answer 1

提示
ACB = 90度
tan角CAB = CB / CA and CB^2 + CA^2 = AB^2 = 10^2 => get CB = ______ and CA = ________
tan角CAD = x => x = tan(0.5 * CAB) => 2x / (1-x^2) = (3 / 4) => 3 * x^2 + 8x – 3 = 0 => x = 1 / 3 or x = -3 (discarded) => cos CAD = 3 / sqrt(10) = CA / CD => CD = sqrt(10) * CA / 3 = 8 * sqrt(10 ) / 3