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 104735 by bemath last updated on 23/Jul/20

Commented by 1549442205PVT last updated on 23/Jul/20

$$\mathbf{what}\mathbf{is}\mathbf{your}\mathbf{question}?$$

Commented by bemath last updated on 23/Jul/20

$$findthevalueof\sin 18°in$$$$geometrically$$

Answered by 1549442205PVT last updated on 24/Jul/20

Commented by 1549442205PVT last updated on 24/Jul/20

$$\mathrm{Drawing}\mathrm{an}\mathrm{isosceles}\mathrm{triamgle}\mathrm{ABC},\mathrm{CA}=\mathrm{CB}=\mathrm{a}$$$$\mathrm{with}\widehat{\mathrm{ACB}}=36°.\mathrm{Then}\widehat{\mathrm{CAB}}=\widehat{\mathrm{CBA}}=72°.$$$$\mathrm{A}\mathrm{point}\mathrm{D}\mathrm{taken}\mathrm{on}\mathrm{the}\mathrm{side}\mathrm{BC}\mathrm{such}\mathrm{that}$$$$\mathrm{DA}=\mathrm{DC}=\mathrm{m}.\mathrm{Denote}\mathrm{the}\mathrm{midpoont}\mathrm{of}$$$$\mathrm{side}\mathrm{AB}\mathrm{by}\mathrm{E}.\mathrm{Then}\mathrm{we}\mathrm{get}\widehat{\mathrm{DCA}}=\widehat{\mathrm{DAC}}=36°$$$$\Rightarrow \widehat{\mathrm{DAB}}=72°-36°=36°\Rightarrow \widehat{\mathrm{ADB}}=72°,\mathrm{so}$$$$\mathrm{\Delta }\mathrm{ABD}\mathrm{is}\mathrm{isosceles}\mathrm{at}\mathrm{A}\Rightarrow \mathrm{AB}=\mathrm{AD}=\mathrm{m}$$$$\mathrm{and}\mathrm{hence}\mathrm{\Delta }\mathrm{ABD}\backsimeq \mathrm{\Delta }\mathrm{CAB}(\mathrm{a}.\mathrm{a})$$$$\Rightarrow \frac{\mathrm{AD}}{\mathrm{BC}}=\frac{\mathrm{BD}}{\mathrm{AB}}\iff \frac{\mathrm{m}}{\mathrm{a}}=\frac{\mathrm{a}-\mathrm{m}}{\mathrm{m}}\iff {\mathrm{a}}^2-\mathrm{am}-{\mathrm{m}}^2=0$$$$\mathrm{This}\mathrm{is}\mathrm{a}\mathrm{quadratic}\mathrm{equation}\mathrm{for}\mathrm{a}\mathrm{with}$$$$\sqrt{\mathrm{\Delta }}=\sqrt{{\mathrm{m}}^2+{4\mathrm{m}}^2}=\mathrm{m}\sqrt{5}.\mathrm{Hence}$$$$\mathrm{a}=\frac{\mathrm{m}+\mathrm{m}\sqrt{5}}{2}=\frac{\mathrm{m}(\sqrt{5}+1)}{2}.$$$$\mathrm{Since}\mathrm{CE}\mathrm{is}\mathrm{the}\mathrm{bisector}\mathrm{line}\mathrm{of}\mathrm{\Delta }\mathrm{ABC}$$$$,\widehat{\mathrm{BCE}}=\frac{1}{2}\widehat{\mathrm{ACB}}=18°\mathrm{and}\mathrm{CE}\mathrm{\perp }\mathrm{AB},\mathrm{so}$$$$\sin 18°=\sin \widehat{\mathrm{BCE}}=\frac{\mathrm{BE}}{\mathrm{BC}}=\frac{\mathrm{m}}{2}:\mathrm{a}=\frac{\mathrm{m}}{2}:\frac{\mathrm{m}(\sqrt{5}+1)}{2}$$$$=\frac{1}{\sqrt{5}+1}=\frac{1\times (\sqrt{5}-1)}{\left(\sqrt{5+1}\right)(\sqrt{5}-1)}=\frac{\sqrt{5}-1}{4}$$$$\mathbf{Thus},\mathbf{sin}18°=\frac{\sqrt{5}-1}{4}\mathbf{which}\mathbf{we}\mathbf{need}\mathbf{to}\mathbf{find}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com