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​

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​

in progress 0
Nevaeh 13 hours 2021-10-12T14:49:49+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-12T14:51:24+00:00

    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)}

    ★★★★★★★★★★★★★★★★★★★

    “And I… am…. Iron Man” (*snaps*)

    <img src="https://tenor.com/view/iron-man-tony-stark-avengers-endgame-iam-iron-man-gif-14595808.gif" height=220 width=330>

    0
    2021-10-12T14:51:24+00:00

    Answer:

    r2sin2Acos2B+r2sin2Asin2B+z2=r2

    r2sin2A(1)+z2=r2

    z2=r2(1-sin2A)

    z2=r2cos2A

Leave an answer

Browse

14:4+1-6*5-7*14:3+5 = ? ( )