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 113667 by ZiYangLee last updated on 14/Sep/20

$$\mathrm{In}△\mathrm{ABC},\mathrm{BC}=5\mathrm{cm}\mathrm{AC}=4\mathrm{cm}$$$$\cos (\mathrm{A}-\mathrm{B})=\frac{31}{32}$$$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}△\mathrm{ABC}.$$

Answered by som(math1967) last updated on 14/Sep/20

$${\tan }^2\frac{\mathrm{A}-\mathrm{B}}{2}=\frac{1-\cos (\mathrm{A}-\mathrm{B})}{1+\cos (\mathrm{A}-\mathrm{B})}$$$$\tan \frac{\mathrm{A}-\mathrm{B}}{2}=\sqrt{\frac{1}{63}}=\frac{1}{3\sqrt{7}}$$$$\frac{5-4}{5+4}\times \cot \frac{\mathrm{C}}{2}=\frac{1}{3\sqrt{7}}★$$$$\cot \frac{\mathrm{C}}{2}=\frac{3}{\sqrt{7}}$$$$\therefore \sin \frac{\mathrm{C}}{2}=\frac{\sqrt{7}}{4},\cos \frac{\mathrm{C}}{2}=\frac{3}{4}$$$$\therefore \mathrm{sinC}=2\times \frac{\sqrt{7}}{4}\times \frac{3}{4}=\frac{3\sqrt{7}}{8}$$$$\mathrm{area}\mathrm{of}\mathrm{triangle}$$$$=\frac{1}{2}\times 5\times 4\times \frac{3\sqrt{7}}{8}$$$$\frac{15\sqrt{7}}{4}{\mathrm{cm}}^2$$$$★\tan \frac{\mathrm{A}-\mathrm{B}}{2}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\times \cot \frac{\mathrm{C}}{2}$$

Commented by ZiYangLee last updated on 16/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}★$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com