va rog am nevoie foarte urgent de acest ex rog seriozitate te rog mult targoviste44 rezolva tu acest ex .

Răspuns:
Explicație pas cu pas:
a) tgB = CE/EB =3/4
20/EB = 3/4, EB = 20*3/4 = 15
CB^2 = 20^2 +15^2 = 625,
CB = 25
PerimABCD = 3*20 +15 +25 = 100
b) sin(arcsinx) = x, cos(arccosx) = x
sinACB = sin(ACE +ECB) = sin(45 +ECB)
sinECB = EB/CB = 15/25 = 3/5
ECB = arcsin(3/5)
sinACB = sin(45 +arcsin(3/5)) =
sin45*cos(arcsin(3/5)) + cos45*sin(arcsin(3/5))
Dar arcsin(3/5) = arccos(√(1 -(3/5)^2)=
=arccos√(16/25)= arccos(4/5)
sinACB = √2/2*4/5 + √2/2*3/5 =
= 4√2/10 + 3√2/10 = 7√2/10