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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=(1/(sin θ)) tan (θ/2)=(1/2) ⇒sin θ=((2×(1/2))/(1+((1/2))^2 ))=(4/5) ⇒x=(5/4)

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

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

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

tan (θ/2)=((1/2)/1)=(1/2) sin θ=2 sin (θ/2) cos (θ/2)=((2 sin (θ/2))/(cos (θ/2)))×cos^2 (θ/2) =((2 tan (θ/2))/(cos^2 (θ/2)+sin^2 (θ/2)))×cos^2 (θ/2) ⇒sin θ=((2 tan (θ/2))/(1+tan^2 (θ/2))) ←just remember it,it′s in every book. =((2×(1/2))/(1+((1/2))^2 )) =(4/5)

$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}/\mathrm{2}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta=\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}×\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}×\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}\:\:\leftarrow{just}\:{remember}\:{it},{it}'{s}\:{in}\:{every}\:{book}. \\ $$$$=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

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

Commented by Tawa1 last updated on 16/Jul/19

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

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

Commented by Tawa1 last updated on 16/Jul/19

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

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

Commented by Tawa1 last updated on 16/Jul/19

Please sir, i understand how you got x = (1/(sinθ)) from sinθ = (1/x) But i don′t understand how you got: tan((θ/2)) = (1/2) and sinθ = ((2 × (1/2))/(1 + ((1/2))^2 )) ? Thanks sir

$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\:\:\:\mathrm{x}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{sin}\theta}\:\:\:\mathrm{from}\:\:\:\:\mathrm{sin}\theta\:\:=\:\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}:\:\:\:\:\:\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{and}\:\:\:\:\:\:\:\mathrm{sin}\theta\:\:=\:\:\frac{\mathrm{2}\:×\:\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}\:+\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:? \\ $$$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 16/Jul/19

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

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

Answered by MJS last updated on 16/Jul/19

center of circle = ((0),(0) ) line: y=−(1−h)x+((h+1)/2) circle: x^2 +y^2 =(1/4) line∩circle with exactly 1 solution ⇒ x^2 +((h^2 −1)/(h^2 −2h+2))x+((h(h+2))/(4(h^2 −2h+2)))=0 x=t−((h^2 −1)/(2(h^2 −2h+2))) t^2 +((4h−1)/(4(h^2 −2h+2)^2 ))=0 t=0 ⇒ h=(1/4) ⇒ x= determinant ((( (((−(1/2))),(1) ) − (((1/2)),((1/4)) ))))=(5/4)

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

Commented by Tawa1 last updated on 16/Jul/19

I wish to know this too, but i don′t understand anything. i only understand that the centre of circle is 0, 0 Hahahaha

$$\mathrm{I}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{know}\:\mathrm{this}\:\mathrm{too},\:\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{anything}. \\ $$$$\mathrm{i}\:\mathrm{only}\:\mathrm{understand}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{is}\:\:\mathrm{0},\:\mathrm{0} \\ $$$$\mathrm{Hahahaha} \\ $$

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

thanks sir for coordinate method!

$${thanks}\:{sir}\:{for}\:{coordinate}\:{method}! \\ $$

Commented by MJS last updated on 16/Jul/19

for Tawa: a circle with center ((p),(q) ) and radius r has the equation: (x−p)^2 +(y−q)^2 =r^2 or y=q±(√(r^2 −(x−p)^2 )) a line through 2 points ((a),(b) ) and ((c),(d) ) has the equation: (d−b)x+(a−c)y−ad+bc=0 or y=((b−d)/(a−c))x+((ad−bc)/(a−c)) a polynomial of degree 2 x^2 +px+q=0 can be solved like this: put x=t−(p/2) ⇒ t^2 −(p^2 /4)+q=0 ⇒ t^2 =(p^2 /4)−q which leads to the well known formula x=−(p/2)±(√((p^2 /4)−q)) a circle intersected by a line in only one point circle: x^2 +y^2 =r^2 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 +((2ab)/(a^2 +1))x+((b^2 −r^2 )/(a^2 +1))=0 we are interested in exactly one solution ⇒ ⇒ x=−(p/2)±0 ⇒ (p^2 /4)−q=0 (1/4)(((2ab)/(a^2 +1)))^2 −((b^2 −r^2 )/(a^2 +1))=0 if we know the values of all variables but one we can solve this without solving the quadratic

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

Commented by Tawa1 last updated on 16/Jul/19

Wow, i understand now sir. clear. God bless you more sir.

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

Commented by Tawa1 last updated on 16/Jul/19

It means r = (1/2) , how is it: (1/2)

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

Commented by MJS last updated on 16/Jul/19

the radius of the circle fits 2 times into the side length of the square which is 1 btw the line x starts at (((−(1/2))),(1) ) and ends at (((1/2)),(h) ) if we put the center of the circle in ((0),(0) ) ⇒ the only unknown is h the length of x is given by (√((−(1/2)−(1/2))^2 +(1−h)^2 )) because the distance between two points ((a),(b) ) and ((c),(d) ) is (√((a−c)^2 +(b−d)^2 ))

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

Commented by Tawa1 last updated on 16/Jul/19

God bless you sir. clear. I appreciate

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

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 ⇒t=(1/4) ⇒x=1+t=(5/4)

$$\mathrm{1}^{\mathrm{2}} +\left(\mathrm{1}−{t}\right)^{\mathrm{2}} =\left(\mathrm{1}+{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{4}{t} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{1}+{t}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Commented by Tawa1 last updated on 16/Jul/19

Wow, great, another approach. very short. God bless you sir.

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

Commented by Tawa1 last updated on 16/Jul/19

I understand, using pythagoras theorem.

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

Commented by Tawa1 last updated on 17/Jul/19

I appreciate

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

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