Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 212214 by CrispyXYZ last updated on 06/Oct/24

$$\mathrm{Find}\tan \theta .$$

Commented by CrispyXYZ last updated on 06/Oct/24

Answered by a.lgnaoui last updated on 06/Oct/24

$$\sin \mathit{\theta }=\frac{\mathrm{CD}}{\mathrm{BC}}\mathrm{BC}=2\sqrt{2}\mathrm{CD}=2\sqrt{2}\sin \mathit{\theta }$$$${\mathrm{AD}}^2=1+{\mathrm{CD}}^2-2\mathrm{CDcos}(180-\mathit{\theta })$$$$9=1+8\sin {}^2\mathit{\theta }+2\sqrt{2}\sin 2\mathit{\theta }$$$$8=8\sin {}^2\mathit{\theta }+2\sqrt{2}\sin 2\mathit{\theta }$$$$\sin 2\mathit{\theta }=2\sqrt{2}(1-\sin {}^2\mathit{\theta })$$$$\sin \mathit{\theta }\sqrt{1-\sin {}^2\mathit{\theta }}=\sqrt{2}(1-\sin {}^2\mathit{\theta })$$$$\sin \mathit{\theta }=\sqrt{2(1-\sin {}^2\mathit{\theta })}$$$$3\sin {}^2\mathit{\theta }-2=0\Rightarrow \sin \mathit{\theta }=\sqrt{\frac{2}{3}};(\mathit{\theta }<90)$$$$$$$$\mathit{\theta }=54,735$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com