Beautifully explained step by step. Appreciate it. Keep up the good work. I am sure many people are benefitting from your insightful videos such as this. Kudos!
Area of larger triangle: A = ½(b.h+b.h) = ½.a.b.sinα A = ½(16*6+24*8) = 144cm² sinα = 2*144/(16+24) =3/4 =sin(180°-α) Area of red triangle: A = ½.a.b.sinα = ½*6*8*3/4 A = 18 cm² ( Solved √)
BFD=α..EFC=β..risultano le 2 equazioni 1)..16/sin(90-β)=24/sin(90-α)..cioè cosα=(3/2)cosβ..2)(16-6tgα)^2+36=(24-8tgβ)^2+64...dopo il calcolo di α,β (troppo complesso!!!)..A=(1/2)6*8sin(α+β)
Beautifully explained step by step. Appreciate it. Keep up the good work. I am sure many people are benefitting from your insightful videos such as this. Kudos!
Parabéns pelos vídeos e questões! É sempre bom divulgar uma boa matemática! 🎉🎉🎉🎉🎉
Very challenging problem and quite elegant and clever solution, thanks a lot!
Area of larger triangle:
A = ½(b.h+b.h) = ½.a.b.sinα
A = ½(16*6+24*8) = 144cm²
sinα = 2*144/(16+24) =3/4 =sin(180°-α)
Area of red triangle:
A = ½.a.b.sinα = ½*6*8*3/4
A = 18 cm² ( Solved √)
Same as video.
Excelent Math Booster!!!!
√(13-12cosa)=√(13-12cosa)*6/sina sina=6,S∆=24sina=144
BFD=α..EFC=β..risultano le 2 equazioni 1)..16/sin(90-β)=24/sin(90-α)..cioè cosα=(3/2)cosβ..2)(16-6tgα)^2+36=(24-8tgβ)^2+64...dopo il calcolo di α,β (troppo complesso!!!)..A=(1/2)6*8sin(α+β)
Very good
Another reason for practice. The area is 18 units square.
ha ha ha ha ha