Solving with trigonometry: Without loss of generality, let the sides of the square have length 1 unit. Drop a perpendicular from S to BC and label the intersection as point R, creating right ΔBSR and ΔCSR. The interior angles of a regular pentagon have measure 108°. SR bisects
AB=BC=BQ---> El triángulo ABQ es isósceles ---> Si "O" es el centro del pentágono ---> Ángulos: OSB=54º---> SBC=180º-90º-54º=36º---> SBA=90º+36º=126º---> ABQ=SBA-SBC-CBQ=126º-36º-36º=54º---> AQB=QAB=(180º-54º)/2=63º---> PQA=AQB-PQB=63º-36º=27º. Gracias y saludos.
Congrat again. Your demonstrations let us know the relationship among squares, triangles and pentagons and teach us ways to ptopose ways to solve the problems. Thank you.
Uau! Que problema bonito. Parabéns pela escolha. Eu gostei muito de resolvê-lo. Brasil Setembro de 2024. Wow! What a beautiful problem. Congratulations on your choice. I really enjoyed solving it. Brazil September 2024.
The answer is theta=27 degrees. I think that this is something that a mathlete should comprehend in less than 10 minutes. I could be wrong. Also could you please make a playlist for geometry problems that make use of a pentagon???
I think there is only 2 or 3 videos on this channel that make use of pentagon. So can't make playlist. Videos that gets more views, we as a youtubers are indirectly forced to make more videos of the same type.
PS:gli angoli del pentagono sono 108..l=lato del quadrato..lp(lato pentagono)=(l/2)/cos36.. risulta tgθ=(l-((l/2)/cos36)cos18)/((l/2)/cos36+((l/2)/cos36)sin18)...tgθ=(2cos36-cos18)/(1+sin18)..svolgo i calcoli θ=27
Solving with trigonometry: Without loss of generality, let the sides of the square have length 1 unit. Drop a perpendicular from S to BC and label the intersection as point R, creating right ΔBSR and ΔCSR. The interior angles of a regular pentagon have measure 108°. SR bisects
AB=BC=BQ---> El triángulo ABQ es isósceles ---> Si "O" es el centro del pentágono ---> Ángulos: OSB=54º---> SBC=180º-90º-54º=36º---> SBA=90º+36º=126º---> ABQ=SBA-SBC-CBQ=126º-36º-36º=54º---> AQB=QAB=(180º-54º)/2=63º---> PQA=AQB-PQB=63º-36º=27º.
Gracias y saludos.
Congrat again. Your demonstrations let us know the relationship among squares, triangles and pentagons and teach us ways to ptopose ways to solve the problems. Thank you.
Uau! Que problema bonito. Parabéns pela escolha. Eu gostei muito de resolvê-lo. Brasil Setembro de 2024. Wow! What a beautiful problem. Congratulations on your choice. I really enjoyed solving it. Brazil September 2024.
Obrigado!
The answer is theta=27 degrees. I think that this is something that a mathlete should comprehend in less than 10 minutes. I could be wrong. Also could you please make a playlist for geometry problems that make use of a pentagon???
I think there is only 2 or 3 videos on this channel that make use of pentagon. So can't make playlist. Videos that gets more views, we as a youtubers are indirectly forced to make more videos of the same type.
PS:gli angoli del pentagono sono 108..l=lato del quadrato..lp(lato pentagono)=(l/2)/cos36.. risulta tgθ=(l-((l/2)/cos36)cos18)/((l/2)/cos36+((l/2)/cos36)sin18)...tgθ=(2cos36-cos18)/(1+sin18)..svolgo i calcoli θ=27
Grazie a te sto progredendo in trigonometria
@@jeanmarcbonici9525🤙
(4)^2=16 (4)^2=16 (4)^2=16 (4)^2=16.{16+16+16+16}=64 360°ABCDS/64=5.40ABCDS 5.2^20 5.2^5^4 1.2^12^2 1^1^2 (ABCDS ➖ 2ABCDS+1).
Brilliant.
Trigonometry forever. 27
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