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Question Number 17771 by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

$$inX\stackrel{\mathrm{\Delta }}{YZ}:$$$$1)XY=XZ$$$$2)XS=ST=TZ=ZY$$$$findanypossibleanglesinthis$$$$triangle,byusingthistwoconditions.$$

Answered by mrW1 last updated on 10/Jul/17

$$\mathrm{let}\mathrm{x}=\mathrm{\angle }\mathrm{X}$$$$\mathrm{\angle }\mathrm{Z}=(\pi -\mathrm{\angle }\mathrm{X})/2=\pi /2-\mathrm{x}/2$$$$$$$$\mathrm{XS}=\mathrm{ST}=\mathrm{TZ}=\mathrm{ZY}=1$$$$\mathrm{XT}=2\cos \mathrm{x}$$$$\mathrm{XZ}=1+2\cos \mathrm{x}=\frac{1}{2\cos \mathrm{\angle }\mathrm{Z}}=\frac{1}{2\sin \frac{\mathrm{x}}{2}}$$$$\Rightarrow 2\sin \frac{\mathrm{x}}{2}(1+2\cos \mathrm{x})=1$$$$2\sin \frac{\mathrm{x}}{2}(3-4{\sin }^2\frac{\mathrm{x}}{2})=1$$$${8\sin }^2\frac{\mathrm{x}}{2}-6\sin \frac{\mathrm{x}}{2}+1=0$$$$\mathrm{let}\mathrm{t}=\sin \frac{\mathrm{x}}{2}$$$${\mathrm{t}}^3-\frac{3}{4}\mathrm{t}+\frac{1}{8}=0$$$$\Rightarrow \mathrm{t}=\cos 280°=\sin 10°=\sin \frac{\mathrm{x}}{2}$$$$\Rightarrow \frac{\mathrm{x}}{2}=10°$$$$\Rightarrow \mathrm{x}=20°$$$$$$$$\mathrm{note}:\mathrm{other}\mathrm{solutions}\mathrm{not}\mathrm{suitable}:$$$$\mathrm{t}=\cos 40°=\sin 50°$$$$\mathrm{t}=\cos 160°=\sin 290°$$

Commented by ajfour last updated on 11/Jul/17

$$\frac{\sin 2\theta }{\mathrm{b}}=\frac{\sin \theta }{\mathrm{a}}$$$$\Rightarrow \mathrm{b}=2\mathrm{acos}\theta$$$$\mathrm{Also}\frac{\sin \theta }{\mathrm{a}}=\frac{\sin (\frac{\pi }{2}-\frac{\theta }{2})}{\mathrm{a}+\mathrm{b}}$$$$\Rightarrow (\mathrm{a}+\mathrm{b})\sin \theta =\mathrm{acos}\frac{\theta }{2}$$$$(\mathrm{a}+2\mathrm{acos}\theta )\sin \theta =\mathrm{acos}\frac{\theta }{2}$$$$\Rightarrow \sin \theta +2\sin \theta \cos \theta =\cos \frac{\theta }{2}$$$$\mathrm{if}\mathrm{we}\mathrm{let}\sin \frac{\theta }{2}=\mathrm{t}$$$$2\mathrm{t}+4\mathrm{t}(1-{2\mathrm{t}}^2)=1$$$${8\mathrm{t}}^3-6\mathrm{t}+1=0$$$$\mathrm{same}\mathrm{equation},\mathrm{i}\mathrm{thought}\mathrm{i}\mathrm{was}$$$$\mathrm{getting}\mathrm{another}\mathrm{cubic}..!$$

Commented by ajfour last updated on 11/Jul/17

Commented by b.e.h.i.8.3.417@gmail.com last updated on 11/Jul/17

$$perfectandnice.thanksmaster!$$

Commented by mrW1 last updated on 11/Jul/17

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{trying}\mathrm{from}\mathrm{an}\mathrm{other}$$$$\mathrm{aspect}.$$$$\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{cubic}\mathrm{eqn}.\mathrm{is}\mathrm{in}\mathrm{this}$$$$\mathrm{case}\mathrm{simple},\mathrm{they}\mathrm{are}\cos 40°,\cos 160°$$$$\mathrm{and}\cos 280°.$$$$$$$$\mathrm{but}\mathrm{we}\mathrm{can}\mathrm{use}\mathrm{a}\mathrm{trick}\mathrm{to}\mathrm{make}\mathrm{it}\mathrm{easier}.$$$$\mathrm{let}\alpha =\frac{\theta }{2}$$$$\mathrm{we}\mathrm{have}\mathrm{got}$$$$8{\sin }^3\alpha -6\sin \alpha +1=0$$$$\Rightarrow 1-2(3\sin \alpha -4{\sin }^3\alpha )=0$$$$$$$$\mathrm{on}\mathrm{the}\mathrm{other}\mathrm{side}\mathrm{we}\mathrm{know}$$$$\sin 3\alpha =3\sin \alpha -4{\sin }^3\alpha$$$$$$$$\Rightarrow 1-2\sin 3\alpha =0$$$$\Rightarrow \sin 3\alpha =\frac{1}{2}$$$$\Rightarrow 3\alpha =30°$$$$\Rightarrow \alpha =10°=\frac{\theta }{2}$$$$\Rightarrow \theta =20°$$

Commented by b.e.h.i.8.3.417@gmail.com last updated on 11/Jul/17

$$thanktoallmybestfriends.godblees$$$$youall.$$

Commented by ajfour last updated on 11/Jul/17

$$\mathrm{thanks}\mathrm{for}\mathrm{enlightening}\mathrm{further}\mathrm{sir}.$$$$\mathrm{i}\mathrm{was}\mathrm{wondering}\mathrm{how}\mathrm{the}\mathrm{cubic}$$$$\mathrm{yielded}\theta =20°.$$

Commented by mrW1 last updated on 11/Jul/17

$$\mathrm{I}\mathrm{solved}\mathrm{the}\mathrm{cubic}\mathrm{eqn}.\mathrm{directly}\mathrm{without}$$$$\mathrm{using}\mathrm{the}\mathrm{trick}\mathrm{above}.$$$${\mathrm{t}}^3-\frac{3}{4}\mathrm{t}+\frac{1}{8}=0$$$$\Rightarrow {\mathrm{t}}^3+3\mathrm{pt}+\mathrm{q}=0...\left(\mathrm{i}\right)$$$$\mathrm{with}\mathrm{p}=-\frac{1}{4}\mathrm{and}\mathrm{q}=\frac{1}{8}$$$$\mathrm{the}\mathrm{solution}\mathrm{of}\left(\mathrm{i}\right)\mathrm{is}\mathrm{following}:$$$$\mathrm{D}={\mathrm{q}}^2+{4\mathrm{p}}^3=\frac{1}{8^2}-\frac{4}{4^3}=-\frac{3}{64}<0$$$$\Rightarrow \mathrm{there}\mathrm{are}3\mathrm{real}\mathrm{roots}:$$$${\mathrm{t}}_1=2\sqrt{-\mathrm{p}}\cos \left(\frac{\phi }{3}\right)$$$${\mathrm{t}}_2=2\sqrt{-\mathrm{p}}\cos (\frac{\phi }{3}+120°)$$$${\mathrm{t}}_3=2\sqrt{-\mathrm{p}}\cos (\frac{\phi }{3}+240°)$$$$\mathrm{with}\cos \phi =\frac{-\mathrm{q}}{2\sqrt{-{\mathrm{p}}^3}}=\frac{-1/8}{2\times 1/8}=-\frac{1}{2}$$$$\Rightarrow \phi =120°$$$$\Rightarrow \phi /3=40°$$$$\Rightarrow {\mathrm{t}}_1=2\sqrt{-\mathrm{p}}\cos \left(\frac{\phi }{3}\right)=2\times \frac{1}{2}\times \cos 40°=\cos 40°$$$$\Rightarrow {\mathrm{t}}_2=2\sqrt{-\mathrm{p}}\cos (\frac{\phi }{3}+120°)=\cos 160°$$$$\Rightarrow {\mathrm{t}}_3=2\sqrt{-\mathrm{p}}\cos (\frac{\phi }{3}+240°)=\cos 280°$$$$$$$$\mathrm{but}\mathrm{only}{\mathrm{t}}_3=\cos 280°\mathrm{is}\mathrm{suitable},\mathrm{since}$$$$\cos 280°=\sin 10°$$$$\Rightarrow \mathrm{t}=\sin \frac{\mathrm{x}}{2}=\sin 10°$$$$\Rightarrow \mathrm{x}=20°$$

Commented by ajfour last updated on 11/Jul/17

$$\mathrm{thanks}\mathrm{Sir},\mathrm{let}\mathrm{me}\mathrm{consult}\mathrm{some}$$$$\mathrm{more}\mathrm{literature}\mathrm{on}\mathrm{solution}\mathrm{of}$$$$\mathrm{cubic}\mathrm{equations}.$$

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