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Question Number 64246 by mr W last updated on 16/Jul/19

Answered by mr W last updated on 16/Jul/19

$$x=\frac{1}{\sin \theta }$$$$\tan \frac{\theta }{2}=\frac{1}{2}$$$$\Rightarrow \sin \theta =\frac{2\times \frac{1}{2}}{1+{\left(\frac{1}{2}\right)}^2}=\frac{4}{5}$$$$\Rightarrow x=\frac{5}{4}$$

Commented by mr W last updated on 16/Jul/19

Commented by mr W last updated on 16/Jul/19

$$\tan \frac{\theta }{2}=\frac{1/2}{1}=\frac{1}{2}$$$$\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\frac{2\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\times {\cos }^2\frac{\theta }{2}$$$$=\frac{2\tan \frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}}\times {\cos }^2\frac{\theta }{2}$$$$\Rightarrow \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\leftarrow justrememberit,{it}^{\prime }sineverybook.$$$$=\frac{2\times \frac{1}{2}}{1+{\left(\frac{1}{2}\right)}^2}$$$$=\frac{4}{5}$$

Commented by mr W last updated on 16/Jul/19

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{Sir}\mathrm{MrW},\mathrm{i}\mathrm{will}\mathrm{start}\mathrm{studying}\mathrm{all}\mathrm{this}\mathrm{questions}\mathrm{from}\mathrm{you}\mathrm{now}.$$$$\mathrm{So}\mathrm{when}\mathrm{i}\mathrm{checked}\mathrm{your}\mathrm{solution},\mathrm{i}\mathrm{will}\mathrm{ask}\mathrm{questions}\mathrm{sir}.$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{Sir},\mathrm{does}\mathrm{it}\mathrm{mean}\mathrm{that}\mathrm{every}\mathrm{square}\mathrm{has}\mathrm{a}\mathrm{length}\mathrm{of}1.\mathrm{unless}$$$$\mathrm{specify}.?.\mathrm{And}\mathrm{why}\mathrm{is}\mathrm{length}\mathrm{of}\mathrm{square}1\mathrm{sir}$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{Please}\mathrm{sir},\mathrm{i}\mathrm{understand}\mathrm{how}\mathrm{you}\mathrm{got}\mathrm{x}=\frac{1}{\sin \theta }\mathrm{from}\sin \theta =\frac{1}{\mathrm{x}}$$$$\mathrm{But}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{how}\mathrm{you}\mathrm{got}:\tan \left(\frac{\theta }{2}\right)=\frac{1}{2}$$$$\mathrm{and}\sin \theta =\frac{2\times \frac{1}{2}}{1+{\left(\frac{1}{2}\right)}^2}?$$$$\mathrm{Thanks}\mathrm{sir}$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{Wow},\mathrm{i}\mathrm{understand}\mathrm{very}\mathrm{well}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$$$\mathrm{I}\mathrm{think}\mathrm{i}\mathrm{will}\mathrm{be}\mathrm{learning}\mathrm{each}\mathrm{time}\mathrm{i}\mathrm{saw}\mathrm{questions}\mathrm{on}\mathrm{the}\mathrm{shapes}$$$$\mathrm{Thanks}\mathrm{once}\mathrm{again}\mathrm{sir}.$$

Answered by MJS last updated on 16/Jul/19

$$\mathrm{center}\mathrm{of}\mathrm{circle}=\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$\mathrm{line}:y=-(1-h)x+\frac{h+1}{2}$$$$\mathrm{circle}:x^2+y^2=\frac{1}{4}$$$$\mathrm{line}\cap \mathrm{circle}\mathrm{with}\mathrm{exactly}1\mathrm{solution}$$$$\Rightarrow$$$$x^2+\frac{h^2-1}{h^2-2h+2}x+\frac{h(h+2)}{4(h^2-2h+2)}=0$$$$x=t-\frac{h^2-1}{2(h^2-2h+2)}$$$$t^2+\frac{4h-1}{4{(h^2-2h+2)}^2}=0$$$$t=0\Rightarrow h=\frac{1}{4}$$$$\Rightarrow x=\left|\begin{array}{c}\left(\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right)-\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{4}\end{array}\right)\end{array}\right|=\frac{5}{4}$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{I}\mathrm{wish}\mathrm{to}\mathrm{know}\mathrm{this}\mathrm{too},\mathrm{but}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{anything}.$$$$\mathrm{i}\mathrm{only}\mathrm{understand}\mathrm{that}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{circle}\mathrm{is}0,0$$$$\mathrm{Hahahaha}$$

Commented by mr W last updated on 16/Jul/19

$$thankssirforcoordinatemethod!$$

Commented by MJS last updated on 16/Jul/19

$$\mathrm{for}\mathrm{Tawa}:$$$$\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{center}\left(\begin{array}{c}p\\ q\end{array}\right)\mathrm{and}\mathrm{radius}r\mathrm{has}\mathrm{the}$$$$\mathrm{equation}:$$$${(x-p)}^2+{(y-q)}^2=r^2$$$$\mathrm{or}$$$$y=q\pm \sqrt{r^2-{(x-p)}^2}$$$$\mathrm{a}\mathrm{line}\mathrm{through}2\mathrm{points}\left(\begin{array}{c}a\\ b\end{array}\right)\mathrm{and}\left(\begin{array}{c}c\\ d\end{array}\right)\mathrm{has}\mathrm{the}$$$$\mathrm{equation}:$$$$(d-b)x+(a-c)y-ad+bc=0$$$$\mathrm{or}$$$$y=\frac{b-d}{a-c}x+\frac{ad-bc}{a-c}$$$$\mathrm{a}\mathrm{polynomial}\mathrm{of}\mathrm{degree}2$$$$x^2+px+q=0$$$$\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{like}\mathrm{this}:$$$$\mathrm{put}x=t-\frac{p}{2}\Rightarrow$$$$t^2-\frac{p^2}{4}+q=0\Rightarrow t^2=\frac{p^2}{4}-q$$$$\mathrm{which}\mathrm{leads}\mathrm{to}\mathrm{the}\mathrm{well}\mathrm{known}\mathrm{formula}$$$$x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}$$$$\mathrm{a}\mathrm{circle}\mathrm{intersected}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{in}\mathrm{only}\mathrm{one}\mathrm{point}$$$$\mathrm{circle}:x^2+y^2=r^2$$$$\mathrm{line}:y=ax+b$$$$x^2+{(ax+b)}^2-r^2=0$$$$(a^2+1)x^2+2abx+b^2-r^2=0$$$$x^2+\frac{2ab}{a^2+1}x+\frac{b^2-r^2}{a^2+1}=0$$$$\mathrm{we}\mathrm{are}\mathrm{interested}\mathrm{in}\mathrm{exactly}\mathrm{one}\mathrm{solution}\Rightarrow$$$$\Rightarrow x=-\frac{p}{2}\pm 0\Rightarrow \frac{p^2}{4}-q=0$$$$\frac{1}{4}{\left(\frac{2ab}{a^2+1}\right)}^2-\frac{b^2-r^2}{a^2+1}=0$$$$\mathrm{if}\mathrm{we}\mathrm{know}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{all}\mathrm{variables}\mathrm{but}\mathrm{one}$$$$\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{without}\mathrm{solving}\mathrm{the}\mathrm{quadratic}$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{Wow},\mathrm{i}\mathrm{understand}\mathrm{now}\mathrm{sir}.\mathrm{clear}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{more}\mathrm{sir}.$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{It}\mathrm{means}\mathrm{r}=\frac{1}{2},\mathrm{how}\mathrm{is}\mathrm{it}:\frac{1}{2}$$

Commented by MJS last updated on 16/Jul/19

$$\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{fits}2\mathrm{times}\mathrm{into}\mathrm{the}$$$$\mathrm{side}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{which}\mathrm{is}1$$$$\mathrm{btw}\mathrm{the}\mathrm{line}x\mathrm{starts}\mathrm{at}\left(\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right)\mathrm{and}\mathrm{ends}$$$$\mathrm{at}\left(\begin{array}{c}\frac{1}{2}\\ h\end{array}\right)\mathrm{if}\mathrm{we}\mathrm{put}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\mathrm{in}\left(\begin{array}{c}0\\ 0\end{array}\right)\Rightarrow \mathrm{the}\mathrm{only}\mathrm{unknown}\mathrm{is}h$$$$\mathrm{the}\mathrm{length}\mathrm{of}x\mathrm{is}\mathrm{given}\mathrm{by}$$$$\sqrt{{(-\frac{1}{2}-\frac{1}{2})}^2+{(1-h)}^2}$$$$\mathrm{because}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{points}$$$$\left(\begin{array}{c}a\\ b\end{array}\right)\mathrm{and}\left(\begin{array}{c}c\\ d\end{array}\right)\mathrm{is}\sqrt{{(a-c)}^2+{(b-d)}^2}$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{clear}.\mathrm{I}\mathrm{appreciate}$$

Answered by mr W last updated on 16/Jul/19

Commented by mr W last updated on 16/Jul/19

$$1^2+{(1-t)}^2={(1+t)}^2$$$$1+1-2t+t^2=1+2t+t^2$$$$1=4t$$$$\Rightarrow t=\frac{1}{4}$$$$\Rightarrow x=1+t=\frac{5}{4}$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{Wow},\mathrm{great},\mathrm{another}\mathrm{approach}.\mathrm{very}\mathrm{short}.$$$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa1 last updated on 16/Jul/19

$$\mathrm{I}\mathrm{understand},\mathrm{using}\mathrm{pythagoras}\mathrm{theorem}.$$

Commented by Tawa1 last updated on 17/Jul/19

$$\mathrm{I}\mathrm{appreciate}$$

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