solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)
Answer:
36 teachers
Step-by-step explanation:
total students=128+121+135=384
groups of 32=384/8=12
teachers=12x3=36
Kilograms is probably the best answer.
Step-by-step explanation:
Given :
1- cosA = 1/2
or, CosA = 1 -1/2
Therefore ; CosA = 1/2 = b/h
According to the Pythagoras theorem,
P = root under h^2 - b^2
= root under (2)^2 - (1)^2
= root under 4 -1
= root 3
Again,
SinA = P/h
= root 3 / 2