Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 212233 by Davidtim last updated on 07/Oct/24

$${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=?$$

Commented by Davidtim last updated on 07/Oct/24

$$Ikindlyrequestyouforhelping$$

Answered by mr W last updated on 08/Oct/24

$$\alpha ={\sin }^{-1}\frac{12}{13}={\cos }^{-1}\frac{5}{13}$$$$\beta ={\cos }^{-1}\frac{3}{5}={\sin }^{-1}\frac{4}{5}$$$$\gamma ={\tan }^{-1}\frac{63}{16}$$$$\sin (\alpha +\beta )=\frac{12}{13}\times \frac{3}{5}+\frac{5}{13}\times \frac{4}{5}=\frac{56}{65}$$$$\cos (\alpha +\beta )=\frac{5}{13}\times \frac{3}{5}-\frac{12}{13}\times \frac{4}{5}=-\frac{33}{65}$$$$\tan (\alpha +\beta )=-\frac{56}{33}$$$$\tan (\alpha +\beta +\gamma )=\frac{-\frac{56}{33}+\frac{63}{16}}{1+\frac{56}{33}\times \frac{63}{16}}=\frac{7}{24}$$$$\Rightarrow \alpha +\beta +\gamma =\pi +{\tan }^{-1}\frac{7}{24}$$$$\approx 3.425rad\approx 196.260°$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com