Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 181719 by cortano1 last updated on 29/Nov/22

Answered by mr W last updated on 29/Nov/22

$$sayBA=a$$$${BD}^2=4^2+8^2-2\times 4\times 8\cos 120°$$$${BD}^2={\left(4\sqrt{3}\right)}^2+a^2-2\times 4\sqrt{4}\times a\cos 60°$$$${\left(4\sqrt{3}\right)}^2+a^2-4\sqrt{3}\times a=4^2+8^2+4\times 8$$$$a^2-4\sqrt{3}\times a-64=0$$$$a=2(\sqrt{3}+\sqrt{19})$$$${CA}^2=8^2+{\left(4\sqrt{3}\right)}^2+2\times 8\times 4\sqrt{3}\cos \varphi$$$${CA}^2=4^2+{\left(2(\sqrt{3}+\sqrt{19})\right)}^2-2\times 4\times 2(\sqrt{3}+\sqrt{19})\cos \varphi$$$$4^2+{\left(2(\sqrt{3}+\sqrt{19})\right)}^2-16(\sqrt{3}+\sqrt{19})\cos \varphi =8^2+{\left(4\sqrt{3}\right)}^2+64\sqrt{3}\cos \varphi$$$$(\sqrt{3\times 19}-1)=2(5\sqrt{3}+\sqrt{19})\cos \varphi$$$$\cos \varphi =\frac{\sqrt{57}-1}{2(5\sqrt{3}+\sqrt{19})}=\frac{2\sqrt{19}-3\sqrt{3}}{14}$$$$\Rightarrow \varphi ={\cos }^{-1}\frac{2\sqrt{19}-3\sqrt{3}}{14}\approx 75.43°$$

Commented by cortano1 last updated on 01/Dec/22

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

Answered by mr W last updated on 01/Dec/22

$$alternative:$$$${BD}^2=4^2+8^2+2\times 4\times 8\times \cos 60°=112$$$${CA}^2=8^2+{\left(4\sqrt{3}\right)}^2+2\times 8\times 4\sqrt{3}\times \cos \varphi =112+64\sqrt{3}\cos \varphi$$$$\frac{BD}{\sin 60°}=\frac{CA}{\sin \varphi }$$$$\frac{4\times 112}{3}=\frac{112+64\sqrt{3}\cos \varphi }{{\sin }^2\varphi }$$$$28{\cos }^2\varphi +12\sqrt{3}\cos \varphi -7=0$$$$\cos \varphi =\frac{-3\sqrt{3}+2\sqrt{19}}{14}$$$$\Rightarrow \varphi ={\cos }^{-1}\frac{-3\sqrt{3}+2\sqrt{19}}{14}\approx 75.43°$$

Answered by cortano1 last updated on 01/Dec/22

Terms of Service

Privacy Policy

Contact: info@tinkutara.com