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Question Number 189603 by normans last updated on 19/Mar/23

	Answered by a.lgnaoui last updated on 19/Mar/23
	$△ADF: ∡ADF=90−18=72 ((sin 18)/(DF))=((sin 72)/(AF)) (1) △ABF : ∡BAF=48+18=66; ∡ABF=90−66=24 ((sin 24)/(AF))=((sin 66)/(BF)) BF=BD+DF=6+DF ((sin 24)/(AF))=((sin 66)/(6+DF)) (2) (1) (2)⇒ { ((DF=((AFsin 18)/(sin 72)) (3))),((6sin 24+DFsin 24=AFsin 66)) :} DF=((AFsin 66−6sin 66)/(sin 24)) (4) AFsin 18sin 24=AFsin 72sin 66−6sin 72sin 66 6sin 72sin 66=AF(sin 72sin 66−sin 18sin 24) AF=((6sin 72sin 66)/(sin 72sin 66−sin 18sin 24)) AF=(/) ⇒DF=(/) AD=DFsin 18= (5) △(FCD) ∡CDF=90−x CD^2 =FC^2 +FD^2 =FD^2 +CD^2 cos^2 x CD=((FD)/(sin x)) △ACD ((sin x)/(AD)) =((sin (18+x))/(AC)) ((sin x)/(DFsin 18))=((sin (18+x))/(AF+((DF)/(tan x)))) AFsin X+DFcos X=DFsin 18(sin 18cos x+cos 18sin x) sin x(AF−DFsin 18cos 18)=cos x(DFsin^2 18−DF) tan x=((DF(sin^2 18−1))/(AF−DFsin 18cos 18)) tan x= [((DFsin 28)/(AF(1−((DF)/(2AF))sin 36))] ((DF)/(AF))=((sin 18)/(sin 72)) =((DF)/(AF))×((−cos^2 18)/((2AF−DFsin 36)))×2AF =((2DFcos^2 18 )/(DFsin 36−2AF)) tan x=((2DFcos^2 18)/(DF(sin 36−((2AF)/(DF))))) =((2cos^2 18)/((sin 36−2((sin 72)/(sin 18))))) tan x =0,3249196... x=18°$
	$△ A D F : ∡ A D F = 90 - 18 = 72$ $\frac{\sin 18}{DF} = \frac{\sin 72}{AF} (1)$ $△ A B F : ∡ B A F = 48 + 18 = 66; ∡ A B F = 90 - 66 = 24$ $\frac{\sin 24}{AF} = \frac{\sin 66}{BF} BF = BD + DF = 6 + DF$ $\frac{\sin 24}{AF} = \frac{\sin 66}{6 + DF} (2)$ $(1) (2) \Rightarrow {\begin{cases} DF = \frac{AFsin 18}{\sin 72} (3) \\ 6 \sin 24 + DFsin 24 = AFsin 66 \end{cases}$ $DF = \frac{AFsin 66 - 6 \sin 66}{\sin 24} (4)$ $AFsin 18 \sin 24 = AFsin 72 \sin 66 - 6 \sin 72 \sin 66$ $6 \sin 72 \sin 66 = AF (\sin 72 \sin 66 - \sin 18 \sin 24)$ $AF = \frac{6 \sin 72 \sin 66}{\sin 72 \sin 66 - \sin 18 \sin 24}$ $AF = \frac{}{} \Rightarrow D F = \frac{}{}$ $AD = DFsin 18 = (5)$ $△ (F C D) ∡ C D F = 90 - x$ ${C D}^{2} = {F C}^{2} + {F D}^{2} = {F D}^{2} + {C D}^{2} \cos^{2} x$ $C D = \frac{F D}{\sin x}$ $△ A C D \frac{\sin x}{AD} = \frac{\sin (18 + x)}{A C}$ $\frac{\sin x}{D F \sin 18} = \frac{\sin (18 + x)}{A F + \frac{DF}{\tan x}}$ $AFsin X + DFcos X = DFsin 18 (\sin 18 \cos x + \cos 18 \sin x)$ $\sin x (AF - DFsin 18 \cos 18) = \cos x (DFsin^{2} 18 - DF)$ $\tan x = \frac{D F (\sin^{2} 18 - 1)}{A F - D F \sin 18 \cos 18}$ $\tan x = [\frac{D F \sin 28}{A F (1 - \frac{D F}{2 A F} \sin 36}]$ $\frac{D F}{A F} = \frac{\sin 18}{\sin 72}$ $= \frac{D F}{A F} \times \frac{- \cos^{2} 18}{(2 A F - D F \sin 36)} \times 2 A F$ $= \frac{2 D F \cos^{2} 18}{D F \sin 36 - 2 A F}$ $\tan x = \frac{2 D F \cos^{2} 18}{D F (\sin 36 - \frac{2 A F}{D F})}$ $= \frac{2 \cos^{2} 18}{(\sin 36 - 2 \frac{\sin 72}{\sin 18})}$ $\tan x = 0, 3249196 . . .$ $x = 18 °$
	Commented by mr W last updated on 19/Mar/23

	$i f x = 18 °, t h e n {i t}^{'} s s y m m e t r i c, t h e n$ $24 ° = 6 ° \Rightarrow t h e r e f o r e t o t a l l y w r o n g!$
	Commented by a.lgnaoui last updated on 19/Mar/23

	Commented by normans last updated on 20/Mar/23

	$n o$
	Answered by mr W last updated on 19/Mar/23

	$s a y A F = 1$ $D F = \tan 18$ $B F = \tan (48 + 18) = \tan 66$ $F C = B F \times \tan 6 = \tan 66 \times \tan 6$ $\tan x = \frac{D F}{F C} = \frac{\tan 18}{\tan 66 \times \tan 6}$ $\Rightarrow x = \tan^{- 1} (\frac{\tan 18}{\tan 66 \times \tan 6}) = 54 °$
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