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Question Number 81755 by Power last updated on 15/Feb/20

	Answered by mr W last updated on 15/Feb/20

	Commented by mr W last updated on 15/Feb/20

	$B C = 2$ $B D = C D = \frac{1}{\cos β}$ $A B = \frac{1}{\cos 2 β}$ $\frac{B E}{\sin β} = \frac{E C}{\sin 2 β} = \frac{B C}{\sin 3 β}$ $\frac{B E}{\sin β} = \frac{E C}{2 \sin β \cos β} = \frac{2}{\sin β (3 - 4 \sin^{2} β)}$ $B E = \frac{2}{3 - 4 \sin^{2} β}$ $E C = \frac{4 \cos β}{3 - 4 \sin^{2} β} = B F$ $Δ_{B D C} = \frac{2 \times 1 \times \tan β}{2} = \tan β = S$ $Δ_{A B F} = \frac{A B \times B F \times \sin β}{2} = \frac{4 \cos β \sin β}{2 \cos 2 β (3 - 4 \sin^{2} β)} = \frac{\tan 2 β}{3 - 4 \sin^{2} β}$ $Δ_{E B D} = \frac{B E \times B D \times \sin β}{2} = \frac{2 \sin β}{2 (3 - 4 \sin^{2} β) \cos β} = \frac{\tan β}{3 - 4 \sin^{2} β}$ $Δ_{A B F} - Δ_{E B D} = \frac{\tan 2 β - \tan β}{3 - 4 \sin^{2} β} = 2 S = 2 \tan β$ $\frac{\tan 2 β - \tan β}{3 - 4 \sin^{2} β} = 2 \tan β$ $\frac{\frac{2 \tan β}{1 - \tan^{2} β} - \tan β}{3 - 4 \sin^{2} β} = 2 \tan β$ $\frac{2 - 2 \sin^{2} β}{1 - 2 \sin^{2} β} - 1 = 2 (3 - 4 \sin^{2} β)$ $\frac{1}{1 - 2 \sin^{2} β} = 6 - 8 \sin^{2} β$ $1 = (6 - 8 \sin^{2} β) (1 - 2 \sin^{2} β)$ $16 \sin^{4} β - 20 \sin^{2} β + 5 = 0$ $\sin^{2} β = \frac{5 \pm \sqrt{5}}{8}$ $\sin β = \frac{\sqrt{2 (5 \pm \sqrt{5})}}{4}$ $\Rightarrow β = \sin^{- 1} \frac{\sqrt{2 (5 \pm \sqrt{5})}}{4} = 36 ° o r 72 °$ $b u t 2 β < 90 °, \Rightarrow β = 36 °$ $\Rightarrow α = 180 ° - 4 β = 36 °$
	Commented by Power last updated on 15/Feb/20

	$great sir$
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