PM=MN=PN=10 cm
PP'=5
teorema lui Pitagora in triunghiul MPP'
MP'=√(MP²-PP'²)=√(10²-5²)=√(100-25)=√75-5√3
a)sin (∡PMP')=5/10=1/2 ⇒ m(∡PMP')=30°
b)
ducem MA perpendicular pe PN notam cu AA' perpendiculara pe plan
AM=√(MP²-AP²)=√(10²-5²)=√(100-25)=√75=5√3
sin (∡AMA')=5/5√3=1/√3 =√3/3
c)
MA'=√(MP'²-A'P'²)=√[(5√3)²-5²]=√(75-25)=√50=5√2
A(MN'P')=10×5√2/2=25√2 cm²