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Question Number 177978 by mr W last updated on 11/Oct/22

	Answered by mr W last updated on 11/Oct/22

	$m e t h o d 1$ $A B = A F = D C = 1$ $B D = 2 \sin 40 °$ $\frac{\sin (C + 110 °)}{\sin C} = \frac{1}{2 \sin 40 °}$ $\cos 110 ° + \frac{\sin 110 °}{\tan C} = \frac{1}{2 \sin 40 °}$ $- \sin 20 ° + \frac{\cos 20 °}{\tan C} = \frac{1}{2 \sin 40 °}$ $\tan C = \frac{2 \sin 40 ° \cos 20 °}{1 + 2 \sin 40 ° \sin 20 °}$ $\Rightarrow C = 40 °$ $\Rightarrow ? = 360 - 80 - 160 - 40 = 80 °$
	Commented by Tawa11 last updated on 11/Oct/22

	$Great sir$
	Answered by mr W last updated on 11/Oct/22

	Commented by mr W last updated on 11/Oct/22

	$m e t h o d 2$ $l e t A E / / D C, C E / / C D$ $\Rightarrow A D C E i s p a r a l l e l o g r a m$ $\Rightarrow A E = A D = A B$ $∠ B A E = 80 - 20 = 60 °$ $\Rightarrow Δ A B E i s e q u i l a t e r a l$ $\Rightarrow B E = A E = A D = E C$ $\Rightarrow Δ B E C i s i s o s c e l e s$ $\Rightarrow ∠ E B C = \frac{180 - 140}{2} = 20 °$ $\Rightarrow ? = 60 + 20 = 80 ° ✓$
	Commented by Tawa11 last updated on 11/Oct/22

	$Great sir$
	Answered by a.lgnaoui last updated on 11/Oct/22

	$Δ ADC$ $AD = DC \Rightarrow ∡ DAC = ∡ DCA = \frac{180 - 160}{2} = 10$ $BAC = 80 - 10 = 70 160 = 70 + 9 \overset{}{0}$ ${AC}^{2} = {2 AD}^{2} - {2 AD}^{2} \cos 160 = {2 AD}^{2} (1 + \sin 70)$ $AC = AD \sqrt{2 (1 + \sin 70)} = AB \sqrt{2 (1 + \sin 70)} (1)$ $Δ ABC$ ${AC}^{2} = {AB}^{2} + {BC}^{2} - 2 AB \times BCcos x$ $calcul d e BC$ ${BC}^{2} = {AB}^{2} + {AC}^{2} - 2 AB \times ACcos 70$ $= {AB}^{2} + {2 AD}^{2} (1 + \sin 70) - 2 AB (AD \sqrt{2 (1 + \sin 70}) \cos 70$ ${BC}^{2} = {AB}^{2} + {2 AB}^{2} (1 + \sin 70) - {2 AB}^{2} \sqrt{2 (1 + \sin 70}) \cos 70$ $= {AB}^{2} [3 + 2 \sin 70 - 2 \sqrt{2 (1 + \sin 70}) \cos 70]$ $\frac{\sin x}{AC} = \frac{\sin 70}{BC} \Rightarrow \frac{\sin^{2} x}{{AC}^{2}} = \frac{\sin^{2} 70}{{BC}^{2}} \Leftrightarrow \frac{\sin^{2} x}{{2 AB}^{2} (1 + \sin 70)} = \frac{\sin^{2} 70}{{AB}^{2} [3 + 2 \sin 70 - 2 \sqrt{2 (1 + \sin 70)} \cos 70]}$ $\frac{\sin^{2} x}{2 (1 + \sin 70)} = \frac{\sin^{2} 70}{3 + 2 \sin 70 - 2 \sqrt{2 (1 + \sin 70}) \cos 70]} = \frac{0, 88302}{3, 53218}$ $\sin^{2} x = \frac{3, 34480}{3, 53218} = 0, 946955$ $\sin x = \sqrt{0, 946955} = 0, 97311$ $x = \sin^{- 1} (0, 97311) = 76, 68^{°}$
	Commented by mr W last updated on 12/Oct/22

	$y o u {d o n}^{'} t a g r e e t h a t 80 ° i s t h e r i g h t$ $a n s w e r ?$
	Commented by a.lgnaoui last updated on 12/Oct/22

	$p l e a s e t h e e x a c t v a l u e i s 79, 9999 \equiv 80$ $A f t e r v e r i f i c a t i o n c a l c u l$ $\sin^{2} x = \frac{3, 7624059}{3, 8793852} = 0, 96984593$ $\sin x = 0, 98480755 x = 79, 99993 \equiv 80$
	Commented by a.lgnaoui last updated on 12/Oct/22
	$I have considered {_(cos 70=0,342) ^(sin 70=0,9696) but the wrong value is to take more than 4 numbers afrer (virgule)$
	$I h a v e c o n s i d e r e d {_{\cos 70 = 0, 342}^{\sin 70 = 0, 9696}$ $b u t t h e w r o n g v a l u e i s t o t a k e m o r e t h a n 4 n u m b e r s$ $a f r e r (v i r g u l e)$
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