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Question Number 169821 by cortano1 last updated on 10/May/22

Answered by floor(10²Eta[1]) last updated on 10/May/22

$$\mathrm{let}\mathrm{AD}=\mathrm{a}=\mathrm{DC}\mathrm{and}\mathrm{BD}=\mathrm{b}$$$$\mathrm{by}\mathrm{law}\mathrm{of}\mathrm{sines}:$$$$\frac{\mathrm{a}}{\sin 30°}=\frac{\mathrm{b}}{\mathrm{sinx}}$$$$\frac{\mathrm{a}}{\sin 105°}=\frac{\mathrm{b}}{\sin (45°-\mathrm{x})},45°-\mathrm{x}=\mathrm{\angle }\mathrm{BAC}$$$$\Rightarrow \mathrm{a}=\frac{\mathrm{b}}{2\mathrm{sinx}}$$$$\therefore \frac{\frac{\mathrm{b}}{2\mathrm{sinx}}}{\sin (60+45)}=\frac{\mathrm{b}}{\frac{1}{\sqrt{2}}(\mathrm{cosx}-\mathrm{sinx})}$$$$\frac{\mathrm{b}}{\mathrm{sinx}\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)}=\frac{\mathrm{b}\sqrt{2}}{\mathrm{cosx}-\mathrm{sinx}}$$$$\Rightarrow \mathrm{cosx}-\mathrm{sinx}=(\sqrt{3}+1)\mathrm{sinx}$$$$\mathrm{cosx}=(\sqrt{3}+2)\mathrm{sinx}$$$$\Rightarrow \mathrm{tgx}=2-\sqrt{3}\Rightarrow \mathrm{x}=\mathrm{arctg}(2-\sqrt{3})$$

Answered by bobhans last updated on 10/May/22

$$\mathrm{\angle }BAD=45°-x;AD=DC=a$$$$\frac{\sin (45°-x)}{\sin x}=\frac{\sin 105°}{\sin 30°}$$$$\iff \frac{1}{2}\sqrt{2}\cos x-\frac{1}{2}\sqrt{2}\sin x=2\sin 105°\sin x$$$$\iff \frac{1}{2}\sqrt{2}\cos x-\frac{1}{2}\sqrt{2}\sin x=2(\frac{1}{2}\sqrt{3}\frac{1}{2}\sqrt{2}+\frac{1}{2}\frac{1}{2}\sqrt{2})\sin x$$$$\iff \cos x-\sin x=(\sqrt{3}+1)\sin x$$$$(\Rightarrow )\cos x=(\sqrt{3}+2)\sin x$$$$(\Rightarrow )\sin {}^{2}x+\cos {}^{2}x=1$$$$(\Rightarrow )\sin {}^{2}x+(7+4\sqrt{3})\sin {}^{2}x=1$$$$(\Rightarrow )\sin {}^{2}x=\frac{1}{8+4\sqrt{3}}$$$$(\Rightarrow )\sin x=\frac{1}{2}.\sqrt{\frac{1}{2+\sqrt{3}}}=\frac{\sqrt{2-\sqrt{3}}}{2}$$$$(\Rightarrow )x=15°$$

Commented by cortano1 last updated on 10/May/22

$$\frac{\sqrt{2-\sqrt{3}}}{2}=\sqrt{\frac{1}{2}-\sqrt{\frac{3}{16}}}=\sqrt{\frac{\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{3}{16}}}{2}}-\sqrt{\frac{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{3}{16}}}{2}}$$$$=\sqrt{\frac{\frac{1}{2}+\frac{1}{4}}{2}}-\sqrt{\frac{\frac{1}{2}-\frac{1}{4}}{2}}$$$$=\sqrt{\frac{3}{8}}-\sqrt{\frac{1}{8}}=\frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}$$$$=\frac{1}{4}\sqrt{2}(\sqrt{3}-1)=\sin 15°$$

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