The Circle Has A Radius 5cm And the angle between sector from the chord is 73.7397952916880° and their right triangle is AB=4 cm AC=3 cm Find the area of arc triangle EDB with E^⌢ D^⌢ is arc


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Question Number 182178 by BagusSetyoWibowo last updated on 05/Dec/22

$T h e C i r c l e H a s A R a d i u s 5 c m$ $A n d t h e a n g l e b e t w e e n$ $s e c t o r f r o m t h e c h o r d i s$ $73.7397952916880 °$ $a n d t h e i r r i g h t t r i a n g l e i s$ $A B = 4 c m$ $A C = 3 c m$ $F i n d t h e a r e a o f a r c t r i a n g l e$ $E D B$ $w i t h \overset{⌢}{E} \overset{⌢}{D} i s a r c$
Commented by BagusSetyoWibowo last updated on 05/Dec/22

Commented by mr W last updated on 05/Dec/22

$b u t t h e c i r c l e h a s a r a d i u s 3 c m!$ $r a d i u s 5 c m a n d a n g l e 73.7397 . . . ° a r e$ $t o o m u c h i n f o r m a t i o n . t h e y a r e i n$ $c o n t r a d i c t i o n w i t h A C = 3 a n d A B = 4 .$ $t h a t m e a n s y o u o n l y n e e d t o g i v e :$ $A C = 3 c m, A B = 4 c m .$
	Answered by mr W last updated on 05/Dec/22

	Commented by mr W last updated on 05/Dec/22

	$\tan ∠ A C B = \frac{4}{3}$ $2 \tan^{- 1} \frac{4}{3} + \frac{π}{2} - α = π$ $α = 2 \tan^{- 1} \frac{4}{3} - \frac{π}{2}$ $\cos α = \sin (2 \tan^{- 1} \frac{4}{3}) = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}$ $s h a d e d a r e a = \frac{3 \times 4}{2} - \frac{3^{2} \sin (\frac{π}{2} - α)}{2} - \frac{3^{2} \times α}{2}$ $= 6 - \frac{9}{2} (\cos α + α)$ $= 6 - \frac{9}{2} (\frac{24}{25} + 2 \tan^{- 1} \frac{4}{3} - \frac{π}{2})$ $= \frac{42}{25} + \frac{9 π}{4} - 9 \tan^{- 1} \frac{4}{3} \approx 0.403$
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