Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 193511 by Lekhraj last updated on 15/Jun/23

	Answered by a.lgnaoui last updated on 17/Jun/23
	$∡OAE=α OA=OB=OF=R ⇒∡OAF=∡OFA=α ⇒cos α=((AF)/(2R))=((11)/(2R)) ⇒ R=((11)/(2cos 𝛂)) (1) Surface triangle (AOE) est: S=((OA×AEsin 𝛂)/2)=((11×4sin 𝛂)/(cos 𝛂)) S=44tan𝛂 (2) avec { ((AE=16)),((OA=R=)) :} • triangle ( OBF) BF^2 =2R^2 (1−cos ϕ) (ϕ=(π/2) −(π−2α)=2α−(π/2) ) ⇒ BF^2 =2R^2 (1−sin 2𝛂) (3) •triangle (ABE) AE⊥BE (BE=r) AB^2 =AE^2 +r^2 ⇒ 2R^2 =16^2 +r^2 •BEF BF^2 =BE^2 +EF^2 =r^2 +25 =2R^2 −231 (4) 2R^2 (1−sin 2𝛂)=2R^2 −231 2R^2 sin 2𝛂−231=0 2(((11^2 )/(4cos^2 𝛂)))sin 2𝛂=231 (sin 2α=((2t)/(1+t^2 )), (1/(cos^2 𝛂))=1+t^2 ; t=tan α) 121t=231 t=((231)/(121)) (5) dans l expression(2) S=44tan 𝛂=((924)/(11)) Surface Triangle (AOE) =84$
	$∡ OAE = α OA = OB = OF = R$ $\Rightarrow ∡ OAF = ∡ OFA = α$ $\Rightarrow \cos α = \frac{AF}{2 R} = \frac{11}{2 R} \Rightarrow R = \frac{11}{2 \cos α} (1)$ $Surface triangle (AOE) est :$ $S = \frac{OA \times AE \sin α}{2} = \frac{11 \times 4 \sin α}{\cos α}$ $S = 44 \tan α (2)$ $avec {\begin{cases} AE = 16 \\ OA = R = \end{cases}$ $∙ triangle (OBF) {BF}^{2} = {2 R}^{2} (1 - \cos φ)$ $(φ = \frac{π}{2} - (π - 2 α) = 2 α - \frac{π}{2})$ $\Rightarrow {BF}^{2} = 2 R^{2} (1 - \sin 2 α) (3)$ $∙ triangle (ABE) AE ⊥ BE (BE = r)$ ${AB}^{2} = {AE}^{2} + r^{2} \Rightarrow 2 R^{2} = 16^{2} + r^{2}$ $∙ BEF {BF}^{2} = {BE}^{2} + {EF}^{2} = r^{2} + 25$ $= 2 R^{2} - 231 (4)$ $2 R^{2} (1 - \sin 2 α) = 2 R^{2} - 231$ $2 R^{2} \sin 2 α - 231 = 0$ $2 (\frac{11^{2}}{4 \cos^{2} α}) \sin 2 α = 231$ $(\sin 2 α = \frac{2 t}{1 + t^{2}}, \frac{1}{\cos^{2} α} = 1 + t^{2}; t = \tan α)$ $121 t = 231 t = \frac{231}{121} (5)$ $dans l expression (2)$ $S = 44 \tan α = \frac{924}{11}$ $Surface Triangle (AOE) = 84$
	Commented by a.lgnaoui last updated on 17/Jun/23

	Commented by Lekhraj last updated on 26/Jun/23

	$T h a n k a l o t f o r t h i s w o n d e r f u l l s o l u t i o n .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum