In the plane, we have
everywhere. So in the equation of the sphere, we have
which is a circle centered at (2, -10, 0) of radius 4.
In the plane, we have
, which gives
But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the plane,
, so
which is a circle centered at (0, -10, 3) of radius .
Answer:
Step-by-step explanation:
(a) ∠FBM is the complement of ∠FBC, so is ...
∠FBM = 90° -52° = 38°
The measure of arc BF is twice this angle, so is ...
arc BF = 2∠FBM = 2(38°)
arc BF = 76°
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(b) ∠M is half the difference between the measures of arcs BE and BF, so is ...
∠M = (1/2)(138° -76°) = 62°/2
∠M = 31°
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(c) arc FC is the supplement to arc BF, so has measure ...
arc FC = 180° -arc BF = 180° -76° = 104°
∠BGE is half the sum of arcs BE and FC, so is ...
∠BGE = (1/2)(arc BE +arc FC) = (138° +104°)/2
∠BGE = 121°
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(d) ∠MFB is the remaining angle in ∆MFB, so has measure ...
∠MFB = 180° -∠M -∠FBM = 180° -31° -38°
∠MFB = 111°
