Question

If x2 + y2 + z2 = r2, where: x = r sin A cos B. y = r sin A sin B, then z
has one of the following values, as​

Answer 1

Answer:

r2sin2Acos2B+r2sin2Asin2B+z2=r2

r2sin2A(1)+z2=r2

z2=r2(1-sin2A)

z2=r2cos2A

Answer 2

Step-by-step explanation:

${x}^{2} + {y}^{2} + {z}^{2} = {r}^{2}$

And,

x = rSinA.CosB

y = rSinA.SinB

Put the given values if x and y in parent equation.

Thus,

${r}^{2} { \sin(a) }^{2} { \cos(b) }^{2} + {r}^{2} { \sin(a) }^{2} { \sin(b) }^{2} + {z}^{2} = {r}^{2}$

${r}^{2} { \sin(a) }^{2} ( { { \sin(b) }^{2} + { \cos(b) }^{2} }) + {z}^{2} = {r}^{2}$

${z}^{2} = {r}^{2} (1 - { \sin(a) )}^{2}$

${z}^{2} = {r}^{2} { \cos(a) }^{2}$

$\fbox{z = rCos(a)}$

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