Question

求直角三角形角的度数.如图,已知BC= M,DE = N,CD = Q; 求角A度数,

Answer 1

tan(A)=BC/AC=m/(CD+AD)=m/(q+n/sin(A))

q*tan(A)+n/cos(A)=m

q*sin(A)+n=m*cos(A)

m*cos(A)-q*sin(A)=n

sin(B-A)=n/sqrt(m*m+q*q)

A=B-arcsin(n/sqrt(m*m+q*q))

其中B=arctan(m/q)