Question

assume that an angle B is an acute angle and tan(5B-9)= cot(7B-81degrees). find the angle B

Answer 1

\[5B-9 = \tan^{-1} (\frac{1}{\tan (7B-81)})\]
\[5B-9 = \frac{1}{7B-81}\]
\[35B^{2}-468B+728 = 0\]
\[B \approx 1.8 , 11.6\]

Answer 2

thank you so much!!!