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Question Number 181918 by HeferH last updated on 02/Dec/22

	Answered by Acem last updated on 02/Dec/22

	Commented by Acem last updated on 02/Dec/22

	$S i r H e f e r H, s h a r e w i t h m e t o f i n d x c a u s e$ $m y b r a i n g o t o f f!$ $i t m a y b e 50 o r 45$
	Commented by Acem last updated on 02/Dec/22

	$T h e r a t i o o f \frac{x}{80 - x} d e p e n d s o n t h e h e i g h t o f t h e$ $i s o c e l l e s t r a p e z o i d, 2 b a s e s a r e p a r a l l e l s$
	Commented by HeferH last updated on 02/Dec/22

	$C u r r e n t l y, I a l s o c a n n o t f i n d t h e a n s w e r$
	Commented by Acem last updated on 02/Dec/22

	$S o m e t h i n g i s m i s s i n g o r i^{'} m d o h u n g r y$
	Commented by HeferH last updated on 02/Dec/22

	$I t s 50 °, b u t m y d r a w i n g i s v e r y m e s s y$
	Answered by HeferH last updated on 02/Dec/22

	Commented by Acem last updated on 02/Dec/22

	$A C D ∡ = 20 n o t 40 S i r$
	Commented by HeferH last updated on 02/Dec/22

	$I t s a e x t e r n a l c o n s t r u c t i o n$
	Commented by HeferH last updated on 02/Dec/22

	Commented by Acem last updated on 02/Dec/22

	$S t i l l n o t c l e a r, w e l l, 1 ∙ h o w y o u k n e w t h a t$ ${A D D}^{'} = 120 - x$ $2 ∙ p l e a s e, c a n y o u d r a w a n o t h e r g r a p h m o r e c l e a r$
	Commented by HeferH last updated on 02/Dec/22

	$1 . 80 - x + 40 = 120 - x$ $(e x t e r i o r a n g l e t h e o r e m)$ $2 . Y e s, I w i l l w o r k o n t h a t, t h e s o l u t i o n$ $i s k i n d a c r a z y$
	Commented by HeferH last updated on 02/Dec/22

	$b t w, t h o s e r e c t a n g l e s, c i r c l e s a n d c u r v e d$ $l i n e s m e a n e q u a l s e g m e n t s$
	Answered by SEKRET last updated on 02/Dec/22

	$\frac{\sin (10 °) \sin (40 °) \sin (x)}{\sin (20 °) \sin (30 °) \sin (80 ° - x)} = 1$
	Commented by Frix last updated on 02/Dec/22

	$\Rightarrow x = 50 °$
	Answered by Frix last updated on 03/Dec/22

	$Let C D = 1 and simply use$ $\frac{\sin α}{a} = \frac{\sin β}{b} = \frac{\sin γ}{c} in each of the triangles$ $\Rightarrow$ $A B = \frac{\sin 40 °}{\cos 10 °}$ $B C = 2 \sin 40 °$ $C A = \frac{\cos 20 °}{\cos 40 °}$ $A D = \frac{\sin 20 °}{\cos 40 °}$ $B D = 2 \sin 10 °$ $C D = 1$ $x = 50 °$
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