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Question Number 191873 by cherokeesay last updated on 02/May/23

Answered by mehdee42 last updated on 03/May/23

$$let:\mathrm{\angle }RPS=p_1\mathrm{\&}\mathrm{\angle }QPS=p_2$$$$p_1=45-x,p_2=135-3x$$$$\frac{PS}{QS}=\frac{sin3x}{{sinp}_2}=\frac{sinx}{{sinp}_1}\Rightarrow \frac{sin3x}{sin(135-3x)}=\frac{sinx}{sin(45-x)}$$$$\Rightarrow sin3x(cosx-sinx)=sinx(cos3x+sin3x)$$$$sin3xcosx-sinxcos3x=2sinxsin3x$$$$sin2x=2sinxsin3x\Rightarrow sinx(sin3x-cosx)=0$$$$inthetriangle;sinx\ne 0$$$$\Rightarrow sin3x=cosx=sin(90-x)$$$$\Rightarrow 3x=360k+90-x\Rightarrow x=22.5✓$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$

Commented by mehdee42 last updated on 03/May/23

$$thistriangleisverticalatthevertexP.$$

Answered by a.lgnaoui last updated on 03/May/23

$$△\mathbf{PQS}\frac{\sin 45}{\mathbf{PQ}}=\frac{\sin 3\mathbf{x}}{\mathbf{PS}}=\frac{\sin (3\mathbf{x}+45)}{\mathbf{QS}}$$$$\Rightarrow \mathrm{PS}=\frac{\sqrt{2}\mathbf{PQ}\sin 3\mathbf{x}}{1}=\frac{\mathbf{QS}\sin 3\mathbf{x}}{\sin (3\mathbf{x}+45)}$$$$\mathbf{PQ}=\frac{\mathbf{QS}}{\sqrt{2}\sin (3\mathbf{x}+45)}$$$$$$$$△\mathrm{PRS}\measuredangle \mathbf{PSR}=\frac{3\mathit{\pi }}{4}$$$$\frac{\sin \mathrm{x}}{\mathbf{PS}}=\frac{\sin 135}{\mathrm{PR}}=\frac{\sin (45-\mathbf{x}))}{\mathbf{SR}}$$$$\Rightarrow \mathbf{PS}=\frac{\mathbf{PR}\sin \mathbf{x}}{\sin 135}$$$$\bullet \mathbf{PQR}\frac{\sin \mathbf{x}}{\mathbf{PQ}=}=\frac{\sin 3\mathbf{x}}{\mathbf{PR}}=\frac{\sin (\mathit{\pi }-4\mathrm{x})}{\mathbf{QS}+\mathbf{SR}}\left(3\right)$$$$\left(1\right)\frac{\sin 45}{\mathbf{PQ}}=\frac{\sin (3\mathbf{x}+45)}{\mathbf{QS}}$$$$\mathbf{QS}=\frac{\mathbf{PQ}\sin (3\mathbf{x}+45)}{\sin 45}$$$$\left(2\right)\mathbf{SR}=\frac{(45-\mathbf{X})\times \mathbf{PR}}{\sin 135}$$$$\mathbf{QS}+\mathbf{SR}=\frac{2\mathbf{PQ}\sin (3\mathbf{X}+45)}{\sqrt{2}}+\frac{\sin (45-\mathbf{x})\times \mathbf{PR}}{\sin 135}$$$$\left(3\right)\Rightarrow \mathbf{PQ}=\frac{\mathbf{PR}\sin \mathbf{x}}{\sin 3\mathbf{x}}$$$$\Rightarrow [\frac{\sin \mathbf{x}\times \sin (3\mathbf{x}+45)}{\sqrt{2}}+\frac{\sin (45-\mathbf{x})}{\sin 135}]\mathbf{PR}=\mathbf{QS}+\mathbf{SR}$$$$$$$$\left(3\right)\Rightarrow \frac{\sin 3\mathrm{x}}{\mathbf{PR}}=\left[\frac{\sqrt{2}\sin 135\sin 4\mathbf{x}}{\sqrt{2}\sin (45-\mathbf{x})+\sin 135\times \sin \mathbf{x}\sin (3\mathbf{x}+45)\mathbf{PR}}\right]$$$$\sin 3\mathbf{x}(\sqrt{2}\sin (45-\mathbf{x})+\sin 135\sin (3\mathbf{x}+45)\sin \mathrm{x}]=$$$$\sqrt{2}\sin 135\times \sin 4\mathbf{x}$$$$\mathrm{to}\mathrm{continous}...........$$$$\sin 3\mathrm{x}[\sqrt{2}(\frac{\sqrt{2}}{2}\cos \mathrm{x}-\frac{\sqrt{2}}{2}\sin \mathrm{x})+\frac{\sqrt{2}}{2}\sin \mathrm{x}(\frac{\sqrt{2}}{2}\sin 3\mathrm{x}+\frac{\sqrt{2}}{2}\cos 3\mathrm{x})$$$$=\sin 4\mathrm{x}$$$$\sin 3\mathrm{x}(\cos \mathrm{x}-\sin \mathrm{x})+\frac{\sin \mathrm{x}(\sin 3\mathrm{x}+\cos 3\mathrm{x})}{2})$$$$\sin 4\mathrm{x}=\sin (3\mathrm{x}+\mathrm{x})=$$$$\sin 3\mathrm{xcos}\mathrm{x}+\cos 3\mathrm{xsin}\mathrm{x}=\sin 3\mathrm{x}(\cos \mathrm{x}+\frac{\sin \mathrm{x}}{\tan 3\mathrm{x}})$$$$\Rightarrow \cos \mathrm{x}+\frac{\sin \mathrm{x}}{\tan 3\mathrm{x}}=\cos \mathrm{x}-\sin \mathrm{x}+\frac{1}{2}(\sin \mathrm{x}+\frac{\sin \mathrm{x}}{\tan 3\mathrm{x}})$$$$\frac{\sin \mathrm{x}}{\tan 3\mathrm{x}}=\frac{\sin \mathrm{x}}{2\tan 3\mathrm{x}}-\frac{1}{2}\sin \mathrm{x}$$$$\Rightarrow \frac{1}{\tan 3\mathbf{x}}=\frac{1}{2\tan 3\mathbf{x}}-\frac{1}{2}$$$$\frac{1}{2\mathbf{tang}3\mathbf{x}}=-\frac{1}{2}\Rightarrow \frac{1}{\mathbf{tang}3\mathbf{x}}=-1$$$$\mathbf{tang}3\mathbf{x}=\mathbf{tang}(-\frac{\mathit{\pi }}{4}+\mathrm{k}\mathit{\pi })$$$$3\mathrm{x}\ne \frac{\pi }{2}+\mathrm{k}\pi \mathrm{x}\ne \mathrm{k}\pi$$$$\mathbf{x}=\frac{-\mathit{\pi }}{12}+\frac{\mathrm{k}\mathit{\pi }}{3}(\mathrm{k}\in \mathbb{Z})$$$$\mathbf{x}=\{-\frac{\mathit{\pi }}{12};\frac{\pi }{4};\frac{3\pi }{4};\frac{5\pi }{4};\frac{11\pi }{12}\}$$$$\mathrm{retenus}\{\frac{\pi }{4},\frac{3\pi }{4}\frac{11\pi }{12}\}$$

Commented by mr W last updated on 03/May/23

$$atrianglewithnegativeangles?$$$$howdoesthetrianglelooklikewith$$$$x=-\frac{\pi }{12}orwithx=\frac{5\pi }{4}?$$

Commented by maths_plus last updated on 03/May/23

$$\mathrm{waoh}!$$

Commented by a.lgnaoui last updated on 03/May/23

$$\mathrm{for}\mathrm{forme}\mathrm{triangle}\mathrm{given}\mathrm{the}\mathrm{solution}\mathrm{conforme}\mathrm{are}$$$$\mathrm{only}\mathrm{angles}:\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4};\frac{11\pi }{12}.$$

Commented by mr W last updated on 03/May/23

$$thenpleasedrawatrianglewith$$$$anyofthesevalues,e.g.x=\frac{11\pi }{12}.$$

Commented by a.lgnaoui last updated on 03/May/23

Commented by a.lgnaoui last updated on 03/May/23

$$\frac{5\pi }{4}:\mathrm{not}\mathrm{valide}$$

Commented by mr W last updated on 03/May/23

$$x=\frac{11\pi }{12}=165°!$$$$3x=495°!$$

Commented by a.lgnaoui last updated on 03/May/23

$$495-360=135=\frac{3\pi }{4}\mathrm{mod}\left(2\pi \right)$$

Commented by a.lgnaoui last updated on 03/May/23

$$\Rightarrow 495\iff 135°\left[2\pi \right]$$$$\mathrm{we}\mathrm{can}\mathrm{retire}\mathrm{it}\mathrm{from}\mathrm{the}\mathrm{solution}$$$$\mathrm{becsuse}\mathrm{its}\mathrm{in}\mathrm{the}\mathrm{contre}-\mathrm{sens}\mathrm{of}\mathrm{angle}$$$$\mathrm{as}\mathrm{in}\mathrm{the}\mathrm{guven}\mathrm{trisngle}$$$$\mathbf{Remsrque}:\mathrm{if}\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{given}\mathrm{in}$$$$\mathrm{the}\mathrm{contraire}\mathrm{sens}\mathrm{like}\left(\mathrm{joint}\mathrm{picture}\right)$$$$\mathrm{x}\mathrm{take}\mathrm{value}(-\mathrm{x})$$$$$$

Commented by a.lgnaoui last updated on 03/May/23

Commented by mr W last updated on 04/May/23

$$thendrawexactlyatrianglewith$$$$yoursolutionx=\frac{\pi }{4}!$$

Answered by ajfour last updated on 04/May/23

$$QP$$$$y=\left(\frac{3m-m^3}{1-3m^2}\right)(x+1)$$$$y=-x$$$$y=m(1-x)$$$$\Rightarrow m-mx+x=0$$$$x=\frac{m}{m-1}$$$$\frac{m}{1-m}=m\left(\frac{3-m^2}{1-3m^2}\right)(\frac{m}{m-1}+1)$$$$\Rightarrow (3-m^2)(2m-1)=3m^2-1$$$$\Rightarrow 2m^3+2m^2-6m+2=0$$$$\Rightarrow m^3+m^2-3m+1=0$$$$x={\tan }^{-1}m$$

Answered by mr W last updated on 04/May/23

$$sayQS=SR=1$$$$PQ=\frac{\sin 45°}{\sin (45°+3x)}$$$$PR=\frac{\sin 45°}{\sin (45°-x)}$$$$\frac{PQ}{PR}=\frac{\sin x}{\sin 3x}$$$$\frac{\sin 45°}{\sin (45°+3x)}\times \frac{\sin (45-x)}{\sin 45°}=\frac{\sin x}{\sin 3x}$$$$\frac{\cos x-\sin x}{\cos 3x+\sin 3x}=\frac{\sin x}{\sin 3x}$$$$\frac{1-\tan x}{1+\tan 3x}=\frac{\tan x}{\tan 3x}$$$$witht=\tan x$$$$\frac{1-t}{1+\frac{3t-t^3}{1-3t^2}}=\frac{t}{\frac{3t-t^3}{1-3t^2}}$$$$t^3+t^2-3t+1=0$$$$t^3-t^2+2t^2-2t-t+1=0$$$$(t-1)(t^2+2t-1)=0$$$$\Rightarrow t=1\Rightarrow rejected!$$$$\Rightarrow t=-1-\sqrt{2}\Rightarrow rejected!$$$$\Rightarrow t=-1+\sqrt{2}$$$$\Rightarrow x={\tan }^{-1}(\sqrt{2}-1)=22.5°$$

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