2 years ago

Quadrilateral ABCD is inscribed in this circle,

Mathematics

1 answer:

neonofarm [45]2 years ago

3 0

Where’s the picture? i can’t answer it without seeing the measurements

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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˚))}

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Sin(A+B) = sinAcosB + cosAsinB

Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)

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