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Question Number 193749 by Mingma last updated on 19/Jun/23

	Answered by som(math1967) last updated on 19/Jun/23

	Commented by som(math1967) last updated on 19/Jun/23
	$AD=DC=AC=12cm ∴∠ACD=60 ∠ACE=tan^(−1) (5/(12)) ∠ECD=60−tan^(−1) ((5/(12))) CE=(√(5^2 +12^2 ))=13cm Area of △CED =(1/2)×CE×CD×sin(60−tan^(−1) (5/(12))) =(1/2)×13×12×{sin60×cos(tan^(−1) (5/(12))) −cos60×sin(tan^(−1) (5/(12)))} =13×6(((√3)/2)×((12)/(13)) −(1/2)×(5/(13))) =13×6×(1/(26))×(12(√3)−5) =3(12(√3)−5)cm^2$
	$A D = D C = A C = 12 c m$ $∴ ∠ A C D = 60$ $∠ A C E = \tan^{- 1} \frac{5}{12}$ $∠ E C D = 60 - \tan^{- 1} (\frac{5}{12})$ $C E = \sqrt{5^{2} + 12^{2}} = 13 c m$ $A r e a o f △ C E D$ $= \frac{1}{2} \times C E \times C D \times s i n (60 - \tan^{- 1} \frac{5}{12})$ $= \frac{1}{2} \times 13 \times 12 \times {s i n 60 \times c o s (\tan^{- 1} \frac{5}{12})$ $- c o s 60 \times s i n (\tan^{- 1} \frac{5}{12})}$ $= 13 \times 6 (\frac{\sqrt{3}}{2} \times \frac{12}{13} - \frac{1}{2} \times \frac{5}{13})$ $= 13 \times 6 \times \frac{1}{26} \times (12 \sqrt{3} - 5)$ $= 3 (12 \sqrt{3} - 5) {c m}^{2}$
	Commented by Mingma last updated on 19/Jun/23
	Perfect ��
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