看图
解:tanα=√3(1+m), α=arctan[√3(1+m)] √3（tanαtanβ+m）+tanβ=0 √3【√3（1+m）tanβ+m】+tanβ=0 tanβ=-√3m/(4+3m) β=arctan[-√3m/(4+3m)] α+β= arctan[√3(1+m)]+ arctan[-√3m/(4+3m)]
