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Question Number 3584 by prakash jain last updated on 16/Dec/15

Prove that ratio of a regular pentagon diagonal to its side is (((√5)+1)/2).

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{pentagon}\:\mathrm{diagonal} \\ $$$$\mathrm{to}\:\mathrm{its}\:\mathrm{side}\:\mathrm{is}\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}. \\ $$

Commented by Yozzii last updated on 15/Dec/15

Its side is one of the 5 edges? a b c e d diagonals=bd,ad,ec,ae,bc?

$${Its}\:{side}\:{is}\:{one}\:{of}\:{the}\:\mathrm{5}\:\:{edges}? \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}\:\:\:\:\:\:\:\:{b} \\ $$$$\:\:\:\:\:{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d} \\ $$$${diagonals}={bd},{ad},{ec},{ae},{bc}? \\ $$$$ \\ $$

Commented by prakash jain last updated on 15/Dec/15

yes. Also it (((√5)+1)/2)

$${yes}.\:\mathrm{Also}\:\mathrm{it}\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Yozzii last updated on 16/Dec/15

Could a general formula for the value of such a ratio be found in terms of n for a regular n−gon?

$${Could}\:{a}\:{general}\:{formula}\:{for}\:{the} \\ $$$${value}\:{of}\:{such}\:{a}\:{ratio}\:{be}\:{found}\:{in}\:{terms} \\ $$$${of}\:{n}\:{for}\:{a}\:{regular}\:{n}−{gon}? \\ $$

Commented by prakash jain last updated on 16/Dec/15

A formula using sin function is clear. A general algebria formula will require solving n^(th) root of unity as an algebric formula.

$$\mathrm{A}\:\mathrm{formula}\:\mathrm{using}\:\mathrm{sin}\:\mathrm{function}\:\mathrm{is}\:\mathrm{clear}.\: \\ $$$$\mathrm{A}\:\mathrm{general}\:\mathrm{algebria}\:\mathrm{formula}\:\mathrm{will}\:\mathrm{require}\:\mathrm{solving} \\ $$$${n}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{as}\:\mathrm{an}\:\mathrm{algebric}\:\mathrm{formula}. \\ $$

Commented by Rasheed Soomro last updated on 16/Dec/15

What is the diagonal of polygon? If a regular polygon has more than one diagonals of digferent sizes then there may be more than one such ratios for example (D_1 /s),(D_2 /s),...

$${What}\:{is}\:{the}\:{diagonal}\:{of}\:{polygon}? \\ $$$${If}\:\:\:{a}\:{regular}\:{polygon}\:{has}\:\:{more}\:{than}\: \\ $$$${one}\:{diagonals}\:{of}\:{digferent}\:{sizes}\:{then} \\ $$$${there}\:{may}\:{be}\:{more}\:{than}\:{one}\:{such}\:{ratios} \\ $$$${for}\:{example}\:\frac{{D}_{\mathrm{1}} }{{s}},\frac{{D}_{\mathrm{2}} }{{s}},... \\ $$

Commented by prakash jain last updated on 16/Dec/15

Yes. There will diagonals of different length.

$$\mathrm{Yes}.\:\mathrm{There}\:\mathrm{will}\:\mathrm{diagonals}\:\mathrm{of}\:\mathrm{different}\:\mathrm{length}. \\ $$

Answered by Yozzii last updated on 16/Dec/15

All regular n−gons can be inscribed in a circle. PROOF: Consider the n roots of the equation z^n =re^(iθ) where i=(√(−1)),r>0,n∈N+{0},−π<θ≤π. Since e^(iθ) =e^(i(θ+2πt)) (t∈Z), because of the periodicity of the sine and cosine functions which arise from the form e^(iθ) =cosθ+isinθ, ⇒ z^n =re^(i(θ+2tπ)) ⇒z=r^(1/n) e^(i((θ+2tπ)/n)) . The n values of z are obtained for 0≤t≤n−1. Observe that each value of z has the common modulus r^(1/n) . On an Argand diagram, say the n values of z are plotted with lines joining their points to the origin. Suppose one point has argument α_t =((θ+2tπ)/n) and an adjacent point has argument α_(t+1) =((θ+2(t+1)π)/n). ⇒ α_(t+1) −α_t =((θ+2tπ+2π−θ−2tπ)/n)=((2π)/n). The angle between the lines joining consecutive points is a constant (((2π)/n)) independent of t and all of these n angular differences sum to 2π. Thus, the n points are regularly positioned about the origin and they may be joined by lines to form a regular n−gon figure. These n points are placed about the origin and they are at the equal distances (r^(1/n) ) from the origin. These points hence, by definition of a circle as the locus of points at a fixed distance from a fixef point, lie on a circle. Since these n points form the corners of a regular n−gon, it follows that such a figure can be inscribed in a circle. □ So, inscribe a pentagon ABCDE in a circle and let its radius be denoted by r. The angle at the centre O of the pentagon in each of the five sectors is ((2π)/5)≡72°. Denote s as the side length of the pentagon. Draw a perpendicular bisector of one side of the pentagon (e.g AB). In so doing, one of the sectors is symmetrically bisected to form two right−angled triangles, with hypothenuses being the radius r of the circle, and s is halved.

$${All}\:{regular}\:{n}−{gons}\:{can}\:{be}\:{inscribed} \\ $$$${in}\:{a}\:{circle}. \\ $$$$\mathrm{PROOF}: \\ $$$${Consider}\:{the}\:{n}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}^{{n}} ={re}^{{i}\theta} \: \\ $$$${where}\:{i}=\sqrt{−\mathrm{1}},{r}>\mathrm{0},{n}\in\mathbb{N}+\left\{\mathrm{0}\right\},−\pi<\theta\leqslant\pi. \\ $$$${Since}\:{e}^{{i}\theta} ={e}^{{i}\left(\theta+\mathrm{2}\pi{t}\right)} \:\left({t}\in\mathbb{Z}\right),\:{because}\:{of}\:{the}\:{periodicity} \\ $$$${of}\:{the}\:{sine}\:{and}\:{cosine}\:{functions}\:{which} \\ $$$${arise}\:{from}\:{the}\:{form}\:{e}^{{i}\theta} ={cos}\theta+{isin}\theta, \\ $$$$\Rightarrow\:{z}^{{n}} ={re}^{{i}\left(\theta+\mathrm{2}{t}\pi\right)} \Rightarrow{z}={r}^{\mathrm{1}/{n}} {e}^{{i}\frac{\theta+\mathrm{2}{t}\pi}{{n}}} . \\ $$$$ \\ $$$${The}\:{n}\:{values}\:{of}\:{z}\:{are}\:{obtained}\:{for}\:\mathrm{0}\leqslant{t}\leqslant{n}−\mathrm{1}. \\ $$$${Observe}\:{that}\:{each}\:{value}\:{of}\:{z}\:{has}\:{the} \\ $$$${common}\:{modulus}\:{r}^{\mathrm{1}/{n}} .\:{On}\:{an}\:{Argand} \\ $$$${diagram},\:{say}\:{the}\:{n}\:{values}\:{of}\:{z}\:{are}\:{plotted} \\ $$$${with}\:{lines}\:{joining}\:{their}\:{points}\:{to}\:{the} \\ $$$${origin}.\:{Suppose}\:{one}\:{point}\:{has}\:{argument} \\ $$$$\alpha_{{t}} =\frac{\theta+\mathrm{2}{t}\pi}{{n}}\:{and}\:{an}\:{adjacent}\:{point}\:{has}\: \\ $$$${argument}\:\alpha_{{t}+\mathrm{1}} =\frac{\theta+\mathrm{2}\left({t}+\mathrm{1}\right)\pi}{{n}}. \\ $$$$\Rightarrow\:\alpha_{{t}+\mathrm{1}} −\alpha_{{t}} =\frac{\theta+\mathrm{2}{t}\pi+\mathrm{2}\pi−\theta−\mathrm{2}{t}\pi}{{n}}=\frac{\mathrm{2}\pi}{{n}}. \\ $$$${The}\:{angle}\:{between}\:{the}\:{lines}\:{joining} \\ $$$${consecutive}\:{points}\:{is}\:{a}\:{constant}\:\left(\frac{\mathrm{2}\pi}{{n}}\right) \\ $$$${independent}\:{of}\:{t}\:{and}\:{all}\:{of}\:{these}\:{n} \\ $$$${angular}\:{differences}\:{sum}\:{to}\:\mathrm{2}\pi.\:{Thus},\:{the} \\ $$$${n}\:{points}\:{are}\:{regularly}\:{positioned}\:{about}\:{the} \\ $$$${origin}\:{and}\:{they}\:{may}\:{be}\:{joined}\:{by}\:{lines} \\ $$$${to}\:{form}\:{a}\:{regular}\:{n}−{gon}\:{figure}. \\ $$$${These}\:{n}\:{points}\:{are}\:{placed}\:{about}\:{the}\: \\ $$$${origin}\:{and}\:{they}\:{are}\:{at}\:{the}\:{equal}\:{distances}\:\left({r}^{\mathrm{1}/{n}} \right) \\ $$$${from}\:{the}\:{origin}.\:{These}\:{points}\:{hence}, \\ $$$${by}\:{definition}\:{of}\:{a}\:{circle}\:{as}\:{the}\:{locus}\:{of} \\ $$$${points}\:{at}\:{a}\:{fixed}\:{distance}\:{from}\:{a}\:{fixef} \\ $$$${point},\:{lie}\:{on}\:{a}\:{circle}.\:{Since}\:{these}\:{n}\:{points} \\ $$$${form}\:{the}\:{corners}\:{of}\:{a}\:{regular}\:{n}−{gon}, \\ $$$${it}\:{follows}\:{that}\:{such}\:{a}\:{figure}\:{can}\:{be} \\ $$$${inscribed}\:{in}\:{a}\:{circle}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\ $$$$ \\ $$$${So},\:{inscribe}\:{a}\:{pentagon}\:{ABCDE}\:{in}\:{a}\:{circle}\:{and} \\ $$$${let}\:{its}\:{radius}\:{be}\:{denoted}\:{by}\:{r}.\:{The} \\ $$$${angle}\:{at}\:{the}\:{centre}\:{O}\:{of}\:{the}\:{pentagon}\:{in} \\ $$$${each}\:{of}\:{the}\:{five}\:{sectors}\:{is}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\equiv\mathrm{72}°. \\ $$$${Denote}\:{s}\:{as}\:{the}\:{side}\:{length}\:{of}\:{the}\:{pentagon}. \\ $$$${Draw}\:{a}\:{perpendicular}\:{bisector}\:{of}\:{one} \\ $$$${side}\:{of}\:{the}\:{pentagon}\:\left({e}.{g}\:{AB}\right).\:{In}\:{so}\:{doing},\:{one} \\ $$$${of}\:{the}\:{sectors}\:{is}\:{symmetrically}\:{bisected} \\ $$$${to}\:{form}\:{two}\:{right}−{angled}\:{triangles}, \\ $$$${with}\:{hypothenuses}\:{being}\:{the}\:{radius}\:{r}\:{of} \\ $$$${the}\:{circle},\:{and}\:{s}\:{is}\:{halved}.\: \\ $$$$ \\ $$

Commented by Yozzii last updated on 16/Dec/15

Thus, in such a triangle, (s/2)=rsin36^° ⇒r=(s/(2sin36°)) . (∗) Now, draw a line joining two points such that a diagonal D of the pentagon is obtained (e.g AC). Draw a perpendicular bisector of AC so that another right−angled triangle is obtained as before. However, the angle at O in this triangle is 72°. Therefore, we get (D/2)=rsin72^° ⇒r=(D/(2sin72°)) . (∗∗) Equating (∗) and (∗∗) we have (D/(2sin72^° ))=(s/(2sin36°))⇒(D/s)=((sin72^° )/(sin36°))=2cos36°.

$${Thus},\:{in}\:{such}\:{a}\:{triangle}, \\ $$$$\frac{{s}}{\mathrm{2}}={rsin}\mathrm{36}^{°} \Rightarrow{r}=\frac{{s}}{\mathrm{2}{sin}\mathrm{36}°}\:.\:\:\left(\ast\right) \\ $$$${Now},\:{draw}\:{a}\:{line}\:{joining}\:{two}\:{points} \\ $$$${such}\:{that}\:{a}\:{diagonal}\:{D}\:{of}\:{the}\:{pentagon} \\ $$$${is}\:{obtained}\:\left({e}.{g}\:{AC}\right).\:{Draw}\:{a}\:{perpendicular} \\ $$$${bisector}\:{of}\:{AC}\:{so}\:{that}\:{another}\:{right}−{angled} \\ $$$${triangle}\:{is}\:{obtained}\:{as}\:{before}.\:{However}, \\ $$$${the}\:{angle}\:{at}\:{O}\:{in}\:{this}\:{triangle}\:{is}\:\mathrm{72}°. \\ $$$${Therefore},\:{we}\:{get} \\ $$$$\frac{{D}}{\mathrm{2}}={rsin}\mathrm{72}^{°} \Rightarrow{r}=\frac{{D}}{\mathrm{2}{sin}\mathrm{72}°}\:.\:\left(\ast\ast\right) \\ $$$${Equating}\:\left(\ast\right)\:{and}\:\left(\ast\ast\right)\:{we}\:{have} \\ $$$$\frac{{D}}{\mathrm{2}{sin}\mathrm{72}^{°} }=\frac{{s}}{\mathrm{2}{sin}\mathrm{36}°}\Rightarrow\frac{{D}}{{s}}=\frac{{sin}\mathrm{72}^{°} }{{sin}\mathrm{36}°}=\mathrm{2}{cos}\mathrm{36}°. \\ $$$$ \\ $$

Commented by Yozzii last updated on 16/Dec/15

Let f=18°⇒ 5f=90°⇒2f=90°−3f. ∴ sin2f=sin(90°−3f) sin(90°−x)=cosx in general. ∴ sin2f=cos3f=4cos^3 f−3cosf But sin2f=2sinfcosf. ∴ 2sinfcosf=4cos^3 f−3cosf. Now, cosf≠0⇒2sinf=4cos^2 f−3 2sinf=4(1−sin^2 f)−3 2sinf=4−4sin^2 f−3 4sin^2 f+2sinf−1=0 ∴sinf=((−2±(√(4−4×4×(−1))))/8) sinf=((−2±2(√5))/8)=((−1±(√5))/4). 0<f<90°⇒sinf>0⇒sinf≠((−1−(√5))/4). ∴ sinf=((−1+(√5))/4) cos^2 f=1−sin^2 f cos^2 f=1−((((√5)−1)/4))^2 =((16−5−1+2(√5))/(16)) cos^2 f=((10+2(√5))/(16))=((5+(√5))/8) ⇒2cos^2 f=((5+(√5))/4)⇒2cos^2 f−1=((1+(√5))/4). But, cos2f=cos36°=2cos^2 f−1. ∴cos36°=((1+(√5))/4)⇒2cos36°=((1+(√5))/2). Hence (D/s)=((1+(√5))/2) ■

$${Let}\:{f}=\mathrm{18}°\Rightarrow\:\mathrm{5}{f}=\mathrm{90}°\Rightarrow\mathrm{2}{f}=\mathrm{90}°−\mathrm{3}{f}. \\ $$$$\therefore\:{sin}\mathrm{2}{f}={sin}\left(\mathrm{90}°−\mathrm{3}{f}\right) \\ $$$${sin}\left(\mathrm{90}°−{x}\right)={cosx}\:{in}\:{general}. \\ $$$$\therefore\:{sin}\mathrm{2}{f}={cos}\mathrm{3}{f}=\mathrm{4}{cos}^{\mathrm{3}} {f}−\mathrm{3}{cosf} \\ $$$${But}\:{sin}\mathrm{2}{f}=\mathrm{2}{sinfcosf}. \\ $$$$\therefore\:\mathrm{2}{sinfcosf}=\mathrm{4}{cos}^{\mathrm{3}} {f}−\mathrm{3}{cosf}. \\ $$$${Now},\:{cosf}\neq\mathrm{0}\Rightarrow\mathrm{2}{sinf}=\mathrm{4}{cos}^{\mathrm{2}} {f}−\mathrm{3} \\ $$$$\mathrm{2}{sinf}=\mathrm{4}\left(\mathrm{1}−{sin}^{\mathrm{2}} {f}\right)−\mathrm{3} \\ $$$$\mathrm{2}{sinf}=\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} {f}−\mathrm{3} \\ $$$$\mathrm{4}{sin}^{\mathrm{2}} {f}+\mathrm{2}{sinf}−\mathrm{1}=\mathrm{0} \\ $$$$\therefore{sinf}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{4}×\mathrm{4}×\left(−\mathrm{1}\right)}}{\mathrm{8}} \\ $$$${sinf}=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}}. \\ $$$$\mathrm{0}<{f}<\mathrm{90}°\Rightarrow{sinf}>\mathrm{0}\Rightarrow{sinf}\neq\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}. \\ $$$$\therefore\:{sinf}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${cos}^{\mathrm{2}} {f}=\mathrm{1}−{sin}^{\mathrm{2}} {f} \\ $$$${cos}^{\mathrm{2}} {f}=\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{16}−\mathrm{5}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$$${cos}^{\mathrm{2}} {f}=\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} {f}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{4}}\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} {f}−\mathrm{1}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}. \\ $$$${But},\:{cos}\mathrm{2}{f}={cos}\mathrm{36}°=\mathrm{2}{cos}^{\mathrm{2}} {f}−\mathrm{1}. \\ $$$$\therefore{cos}\mathrm{36}°=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\Rightarrow\mathrm{2}{cos}\mathrm{36}°=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}. \\ $$$${Hence}\:\frac{{D}}{{s}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$

Answered by Rasheed Soomro last updated on 16/Dec/15

Let A,B,C are any three consecutive vertics, then AB^(−) ,BC^(−) are sides of pentagon and AC^(−) is its diagonal Now let mAB^(−) =mBC^(−) =s and mAC^(−) =d We have to prove d:s=(((√5)+1)/2):1 or (d/s)=(((√5)+1)/2) We know that m∠ABC=108° By cosine law d=(√(s^2 +s^2 −2(s)(s) cos 108°)) (d/s)=((√(s^2 +s^2 −2(s)(s) cos 108°))/s) =((s(√(2−2cos 108°)))/s)=(√(2−2cos 108°)) =(√(2−2(((1−(√5))/4))))=(√((8−2+2(√5))/4)) =((√(6+2(√5)))/2) Now we have to change the numerator into a+b(√5) form,where a ,b are rationals So let (√(6+2(√5))) =a+b(√5) 6+2(√5)=(a+b(√5))^2 =a^2 +5b^2 +2ab(√5) a^2 +5b^2 =6 ∧ 2ab=2⇒a=(1/b) a^2 +5b^2 =6⇒((1/b))^2 +5b^2 =6 ⇒1+5b^4 =6b^2 ⇒5b^4 −6b^2 +1=0 ⇒b^2 =((6±(√(36−20)))/(10))=((6±4)/(10))=1,(1/5) ⇒b=±1,±(1/(√5)) [discarding as its irrational] b=±1⇒a=±1 a+b(√5)=1+(√5),1−(√5),−1+(√5),−1−(√5) Last three are extraneous(?) ((√(6+2(√5)))/2)=((1+(√5))/2)

$${Let}\:{A},{B},{C}\:{are}\:{any}\:{three}\:{consecutive}\: \\ $$$${vertics},\:{then}\:\overline {{AB}}\:\:,\overline {{BC}}\:\:{are}\:{sides}\:{of}\: \\ $$$${pentagon}\:{and}\:\overline {{AC}}\:{is}\:{its}\:{diagonal} \\ $$$${Now}\:{let}\:{m}\overline {{AB}}={m}\overline {{BC}}={s} \\ $$$${and}\:\:{m}\overline {{AC}}\:={d} \\ $$$${We}\:{have}\:{to}\:{prove}\:{d}:{s}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}:\mathrm{1} \\ $$$${or}\:\frac{{d}}{{s}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$${We}\:{know}\:{that}\:{m}\angle{ABC}=\mathrm{108}° \\ $$$${By}\:{cosine}\:{law}\: \\ $$$$\:\:\:\:\:\:\:\:\:{d}=\sqrt{{s}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{2}\left({s}\right)\left({s}\right)\:\mathrm{cos}\:\mathrm{108}°} \\ $$$$\:\:\:\:\:\:\:\:\frac{{d}}{{s}}=\frac{\sqrt{{s}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{2}\left({s}\right)\left({s}\right)\:\mathrm{cos}\:\mathrm{108}°}}{{s}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{s}\sqrt{\mathrm{2}−\mathrm{2cos}\:\mathrm{108}°}}{{s}}=\sqrt{\mathrm{2}−\mathrm{2cos}\:\mathrm{108}°} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{2}−\mathrm{2}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}\right)}=\sqrt{\frac{\mathrm{8}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$${Now}\:{we}\:{have}\:{to}\:{change}\:{the}\:{numerator}\:{into} \\ $$$${a}+{b}\sqrt{\mathrm{5}}\:\:{form},{where}\:{a}\:,{b}\:{are}\:{rationals} \\ $$$${So}\:\:{let}\:\:\:\:\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}\:\:={a}+{b}\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}=\left({a}+{b}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{5}} \\ $$$${a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} =\mathrm{6}\:\:\wedge\:\:\mathrm{2}{ab}=\mathrm{2}\Rightarrow{a}=\frac{\mathrm{1}}{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} =\mathrm{6}\Rightarrow\left(\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} =\mathrm{6} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{5}{b}^{\mathrm{4}} =\mathrm{6}{b}^{\mathrm{2}} \Rightarrow\mathrm{5}{b}^{\mathrm{4}} −\mathrm{6}{b}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{20}}}{\mathrm{10}}=\frac{\mathrm{6}\pm\mathrm{4}}{\mathrm{10}}=\mathrm{1},\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\Rightarrow{b}=\pm\mathrm{1},\pm\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:\left[{discarding}\:{as}\:{its}\:{irrational}\right] \\ $$$${b}=\pm\mathrm{1}\Rightarrow{a}=\pm\mathrm{1} \\ $$$${a}+{b}\sqrt{\mathrm{5}}=\mathrm{1}+\sqrt{\mathrm{5}},\mathrm{1}−\sqrt{\mathrm{5}},−\mathrm{1}+\sqrt{\mathrm{5}},−\mathrm{1}−\sqrt{\mathrm{5}} \\ $$$${Last}\:{three}\:{are}\:{extraneous}\left(?\right) \\ $$$$\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 16/Dec/15

(d/s)=((√(6+2(√5)))/2)=((1+(√5))/2)

$$\frac{{d}}{{s}}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$

Commented by Rasheed Soomro last updated on 16/Dec/15

THαnkS!

$$\mathcal{TH}\alpha{n}\Bbbk\mathcal{S}! \\ $$

Answered by Rasheed Soomro last updated on 16/Dec/15

Let A,B,C are three consecutive vertices of a pentagon.Then AB and BC are sides and AC is diagonal of pentagon. m∠ABC=108° △ABC is an issoscel triangle in which let AB=BC=s, AC=d m∠ABC=108° ,m∠CAB=m∠BCA=36° By sine law (d/(sin 108°))=(s/(36°))⇒(d/s)=((sin 108°)/(sin 36°)) =((((√(5+(√5))))/(((√(5−(√5)))))×((((√(5+(√5))))/(((√(5+(√5)))))=((5+(√5))/(√(25−5)))=((5+(√5))/(2(√5))) =(5/(2(√5)))+((√5)/(2(√5)))=((√5)/2)+(1/2)= (d/s)=((1+(√5))/2)

$${Let}\:{A},{B},{C}\:{are}\:{three}\:{consecutive}\:{vertices} \\ $$$${of}\:{a}\:{pentagon}.{Then}\:{AB}\:{and}\:{BC}\:{are}\:{sides} \\ $$$${and}\:{AC}\:{is}\:{diagonal}\:{of}\:{pentagon}.\:{m}\angle{ABC}=\mathrm{108}° \\ $$$$\bigtriangleup{ABC}\:{is}\:{an}\:{issoscel}\:{triangle}\:{in}\:{which}\: \\ $$$$\:{let}\:{AB}={BC}={s},\:{AC}={d} \\ $$$${m}\angle{ABC}=\mathrm{108}°\:,{m}\angle{CAB}={m}\angle{BCA}=\mathrm{36}° \\ $$$${By}\:{sine}\:{law} \\ $$$$\:\:\:\:\frac{{d}}{{sin}\:\mathrm{108}°}=\frac{{s}}{\mathrm{36}°}\Rightarrow\frac{{d}}{{s}}=\frac{{sin}\:\mathrm{108}°}{{sin}\:\mathrm{36}°} \\ $$$$\:\:\:\:\:=\frac{\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right.}{\left(\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}\right.}×\frac{\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right.}{\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right.}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\sqrt{\mathrm{25}−\mathrm{5}}}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5}}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}= \\ $$$$\frac{{d}}{{s}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$

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