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 26824 by kaivan.ahmadi last updated on 30/Dec/17

Answered by mrW1 last updated on 30/Dec/17

$$\mathrm{\angle }CBD=60-45=15°$$$$\frac{DB}{\sin \mathrm{\angle }C}=\frac{CD}{\sin \mathrm{\angle }CBD}$$$$\Rightarrow DB=\frac{\sin 45}{\sin 15}\times n=(\sqrt{3}+1)n$$$$AB=\sqrt{{AD}^2+{DB}^2-2\times AD\times DB\times \cos \mathrm{\angle }ADB}$$$$=\sqrt{4n^2+(4+2\sqrt{3})n^2-2\times 2n\times (\sqrt{3}+1)n\times \frac{1}{2}}$$$$=n\sqrt{4+(4+2\sqrt{3})-2\times 2(\sqrt{3}+1)\times \frac{1}{2}}$$$$=n\sqrt{8+2\sqrt{3}-2(\sqrt{3}+1)}$$$$=n\sqrt{6}$$$$\frac{DB}{\sin \mathrm{\angle }A}=\frac{AB}{\sin \measuredangle ADB}$$$$\Rightarrow \sin \mathrm{\angle }A=\frac{\sqrt{3}+1}{\sqrt{6}}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}$$$$\Rightarrow \mathrm{\angle }A=75°$$

Commented by kaivan.ahmadi last updated on 30/Dec/17

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

Answered by mrW1 last updated on 30/Dec/17

$$Anotherway:$$$$let\mathrm{\angle }A=\alpha$$$$\mathrm{\angle }ABD=180-60-\alpha =120-\alpha$$$$\frac{DB}{\sin \alpha }=\frac{2n}{\sin (120-\alpha )}$$$$\Rightarrow \frac{n}{DB}=\frac{\sin (120-\alpha )}{2\sin \alpha }$$$$\frac{DB}{\sin 45}=\frac{n}{\sin 15}$$$$\Rightarrow \frac{n}{DB}=\frac{\sin 15}{\sin 45}$$$$\Rightarrow \frac{\sin (120-\alpha )}{2\sin \alpha }=\frac{\sin 15}{\sin 45}$$$$\Rightarrow \frac{\sin (120-\alpha )}{\sin \alpha }=2\times \frac{\sin 15}{\sin 45}=2\times \frac{\sqrt{6}-\sqrt{2}}{4}\times \sqrt{2}=\sqrt{3}-1$$$$\Rightarrow \frac{\sin 120\cos \alpha -\sin \alpha \cos 120}{\sin \alpha }=\sqrt{3}-1$$$$\Rightarrow \frac{\sqrt{3}}{2}\cot \alpha +\frac{1}{2}=\sqrt{3}-1$$$$\Rightarrow \cot \alpha =\frac{2\sqrt{3}-3}{\sqrt{3}}=2-\sqrt{3}$$$$\Rightarrow \alpha =75°$$

Answered by ajfour last updated on 30/Dec/17

$$letperpendicularfromBonAC$$$$beBF=h$$$$thenCF=h(\mathrm{\angle }C=45°)$$$$DF=h\cot 60°=\frac{h}{\sqrt{3}}$$$$AF=h\cot A$$$$AD=2CD$$$$\Rightarrow AF+DF=2(CF-DF)$$$$orh\cot A+\frac{h}{\sqrt{3}}=2(h-\frac{h}{\sqrt{3}})$$$$\Rightarrow \cot A=2-\sqrt{3}$$$$\tan 75°=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{2}{4-2\sqrt{3}}$$$$\Rightarrow \cot 75°=2-\sqrt{3}$$$$soA=75°.$$

Commented by mrW1 last updated on 30/Dec/17

$$\mathcal{VERY}\mathcal{NICE}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com