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Question Number 218267 by lmcp1203 last updated on 04/Apr/25

Commented by lmcp1203 last updated on 04/Apr/25

$$thanks$$

Answered by vnm last updated on 04/Apr/25

$$$$$$\mathrm{Nothing}\mathrm{has}\mathrm{changed}\mathrm{with}\mathrm{the}$$$$\mathrm{drawings},\mathrm{they}\mathrm{still}{\mathrm{won}}^{\prime }\mathrm{t}$$$$\mathrm{load}.\mathrm{The}\mathrm{dot}\mathrm{on}\mathrm{the}\mathrm{left}\mathrm{is}\mathrm{A},\mathrm{on}$$$$\mathrm{the}\mathrm{top}\mathrm{is}\mathrm{B},\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{is}\mathrm{C}$$$$\mathrm{and}\mathrm{at}\mathrm{the}\mathrm{bottom}\mathrm{in}\mathrm{the}$$$$\mathrm{middle}\mathrm{is}\mathrm{D}.$$$$letAC=1,AD=BC=a$$$$\measuredangle ABC=120°$$$$\frac{AC}{\sin \measuredangle ABC}=\frac{BC}{\sin \measuredangle BAC}$$$$a=BC=AC\frac{\sin \measuredangle BAC}{\sin \measuredangle ABC}=\frac{\sin (60°-2x)}{\sin {120}^{°}}=$$$$\frac{\sqrt{3}\cos 2x-\sin 2x}{\sqrt{3}}$$$$\frac{BC}{\sin \measuredangle BDC}=\frac{CD}{\sin \measuredangle CBD}$$$$\frac{a}{\sin (180°-3x)}=\frac{1-a}{\sin x}$$$$\frac{\frac{\sqrt{3}\cos 2x-\sin 2x}{\sqrt{3}}}{\sin 3x}=\frac{1-\frac{\sqrt{3}\cos 2x-\sin 2x}{\sqrt{3}}}{\sin x}$$$$$$$$2x=t$$$$\sqrt{3}({2\cos }^{2}t-1)=2\sin t\cdot (1+\cos t)$$$${16\cos }^{4}t+{8\cos }^{3}t-{12\cos }^{2}t-8\cos t-1=0$$$$2\cos t=u$$$$u^4+u^3-3u^2-4u-1=0$$$$u^3(u+1)-(3u+1)(u+1)=0$$$$(u^3-3u-1)(u+1)=0$$$$u^3-3u-1=0$$$$\frac{1}{2}=\cos 60°=\cos (3\cdot 20°)=$$$$\cos 40°\cos 20°-\sin {40}^{°}\sin 20°=$$$$({2\cos }^{2}20°-1)\cos 20°-$$$$2\cos {20}^{°}(1-{\cos }^{2}20°)=$$$${4\cos }^{3}20°-3\cos {20}^{°}$$$${\left(2\cos 20°\right)}^3-3\left(2\cos 20°\right)-1=0$$$$x=10°$$

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