Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 68576 by naka3546 last updated on 13/Sep/19

Commented by naka3546 last updated on 13/Sep/19

$F i n d a, i f [y e l l o w] = [b l u e]$ $A r e a o f y e l l o w = a r e a o f b l u e$
	Answered by mr W last updated on 14/Sep/19

	$l e t R = 1$ $\frac{1^{2}}{2} [\frac{π}{2} + 2 a - \sin (\frac{π}{2} + 2 a)] - \frac{1^{2}}{2} (\frac{π}{2} - \sin \frac{π}{2}) - \frac{\sqrt{2}}{2} \times \sqrt{2} \tan a = \frac{1}{2} \times \frac{1^{2}}{2} (\frac{π}{2} - \sin \frac{π}{2})$ $\Rightarrow 2 a - \cos 2 a - 2 \tan a = \frac{π}{4} - \frac{3}{2}$ $\Rightarrow \cos^{2} a + \tan a - a = \frac{10 - π}{8}$ $\Rightarrow a \approx 0.4255 = 24.38 °$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum