Question

ajutatima ,multumesc mult!

Answer 1

5)
a)m(<ABD)=60⁰ ⇒ m(<ADB)=30⁰
sin 60⁰ = AD/DB 
√3/2=9/DB
DB=18/√3=6√3
m(<AOB)=m(<COD)=60⁰
m(<DOA)=m(<COB)=120⁰
AB=√(6√3)²-9²=√27=3√3
A=9x3√3=27√3 cm²
P=9x2+3√3x2=18+6√3≈28,39 cm

6.in romb daca m(ADC)=120⁰  ⇒ m(<BAD)=60⁰
rezulta ca ΔBAD=BDC sunt echilaterale deci AB=AD=BC=CD=12√2
P=12√2x4=48√2 cm
AC=2x√[(12√2)²-(6√2)²]=2x√(288-72)=2x √216=2x6√6=12√6

7)
sin 20 =sin(90-70)= sin90 cos70 - sin 70 cos90 = 1 • cos 70-sin 70 • 0 = cos 70
cos 17=cos(90-73)=cos 90cos 73+sin 90sin73=sin 73