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Question Number 177432 by HeferH last updated on 05/Oct/22

Commented by Ar Brandon last updated on 05/Oct/22

$30 ° ?$
Commented by HeferH last updated on 05/Oct/22

$I^{'} m n o t s u r e$
Commented by Ar Brandon last updated on 05/Oct/22

Commented by HeferH last updated on 05/Oct/22

Commented by HeferH last updated on 05/Oct/22

$x + α + 140 ° = 180$ $x = α + 20 °$ $x = 30 °$
Commented by Tawa11 last updated on 05/Oct/22

$Great sirs$
Commented by mr W last updated on 05/Oct/22

$h o w c a n y o u s a y α = 10 ° ? w h e r e i s$ $A E = C D ?$
Commented by HeferH last updated on 05/Oct/22

${I t}^{'} s o n e o f t h o s e m a g i c l i n e s$
Commented by mr W last updated on 05/Oct/22

${w h a t}^{'} s t h e p r o o f f o r i t s c o r r e c t n e s s ?$
Commented by HeferH last updated on 05/Oct/22

$I k n o w {i t}^{'} s a n i n s e c u r e w a y, b u t I s e e s o$ $m a n y t e a c h e r s d r a w c e r t a i n l i n e s t h a t$ $a t f i r s t a r e v a l i d b u t t h e n I t h i n k ‘ ‘ {T h a t}^{'} s$ $n o t p o s s i b l e, i s i t ?^{″}, i n t h i s c a s e I^{'} m$ $t r u s t i n g t h a t b y f o r m i n g 140 ° w i t h t h e$ $d r a w n s e g m e n t t h e e n d t e r m i n a t e s a t$ $a p o i n t F o n A C s u c h t h a t A F = F D = D C$ $P r o b a b l y I h a d l u c k$
	Answered by Ar Brandon last updated on 05/Oct/22

	$(i) \frac{EB}{\sin 40 °} = \frac{h}{\sin 80 °} \Rightarrow EB = h \frac{\sin 40 °}{\sin 80 °} = \frac{h}{2 \cos 40 °}$ $(i i) x + EB = h \Rightarrow x = (1 - \frac{1}{2 \cos 40 °}) h = (\frac{2 \cos 40 ° - 1}{2 \cos 40 °}) h$ $(i i i) \frac{CE}{\sin 60 °} = \frac{h}{\sin 80 °} \Rightarrow CE = h \frac{\sin 60 °}{\sin 80 °}$ $(i v) x + DE = CE \Rightarrow DE = CE - x = (\frac{\sin 60 °}{\sin 80 °} - \frac{2 \cos 40 ° - 1}{2 \cos 40 °}) h$ $Applying cosine rule in △ ADE$ $AD = \sqrt{x^{2} + {DE}^{2} - 2 x . DE . \cos 100 °}$ $= h \sqrt{{(\frac{2 \cos 40 ° - 1}{2 \cos 40 °})}^{2} + {(\frac{\sin 60 °}{\sin 80 °} - \frac{2 \cos 40 ° - 1}{2 \cos 40 °})}^{2} - 2 (\frac{2 \cos 40 ° - 1}{2 \cos 40 °}) (\frac{\sin 60 °}{\sin 80 °} - \frac{2 \cos 40 ° - 1}{2 \cos 40 °}) \cos 100 °}$ $\approx h \sqrt{0, 347296355^{2} + 0, 532088886^{2} - 2 (0, 347296355) (0, 532088886) \cos 100 °}$ $\approx h \sqrt{0, 467911113} = 0, 684040286 h$ $Applying Sine rule we have$ $\frac{\sin ϑ}{x} = \frac{\sin 100 °}{AD} \Rightarrow ϑ = \sin^{- 1} (\frac{x}{AD} \sin 100 °) = \sin^{- 1} (\frac{h}{AD} \cdot \frac{2 \cos 40 ° - 1}{2 \cos 40 °} \cdot \sin 100 °)$ $ϑ = \sin^{- 1} (\frac{(2 \cos 40 ° - 1) (\sin 100 °)}{(0, 684040286) (2 \cos 40 °)}) = 30 °$
	Answered by mr W last updated on 05/Oct/22

	$\frac{A D}{\sin 100} = \frac{A E}{\sin x} . . . (i)$ $\frac{\sin 20}{A D} = \frac{\sin (x - 20)}{C D} . . . (i i)$ $(i) \times (i i) :$ $\frac{\sin 20}{\sin 100} = \frac{\sin (x - 20)}{\sin x}$ $\frac{1}{\cos 10} = \frac{1}{\tan 20} - \frac{1}{\tan x}$ $\tan x = \frac{1}{\frac{1}{\tan 20} - \frac{1}{\cos 10}}$ $\Rightarrow x = \tan^{- 1} (\frac{1}{\frac{1}{\tan 20} - \frac{1}{\cos 10}}) = 30 °$
	Commented by Ar Brandon last updated on 05/Oct/22
	Nice Sir !
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