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Alisiya [41]
3 years ago
9

PLEASE HELP

Mathematics
1 answer:
miss Akunina [59]3 years ago
6 0

In the smallest of the three triangles (with sides a,b,3) we have

a^2+3^2=b^2

In the second-largest triangle (with sides a,108), the length of the missing side is \sqrt{a^2+108^2}. Then in the largest triangle (with sides b,111,\sqrt{a^2+108^2}), we have

b^2+\left(\sqrt{a^2+108^2}\right)^2=(108+3)^2=111^2

\implies b^2+a^2+108^2=111^2

\implies b^2+a^2=657

Equivalently, we have

(a^2+9)+a^2=657\implies 2a^2=648\implies a^2=324\implies a=18

In turn, we find

b^2=18^2+3^2=298\implies b=\sqrt{333}=3\sqrt{37}

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Answer:

To this question; we want to show that if two antipodal points on a sphere were A and B, for any random point C on the sphere, AC is perpendicular to BC?

Step-by-step explanation:

I would proceed by imagining the sphere in a three dimensional Cartesian coordinate system. For example, use a sphere of diameter 2 and let it sit at the origin. Then (0, 0, 1) and (0, 0, -1) are the vector locations of the “north and south poles” of the sphere.

Now choose the vector location of any point on the surface of the sphere. It will have a vector location - we’ll call it (x, y, z). Now the vectors from your point to the two poles are (-x, -y, 1-z) and (-x, -y, -1-z).

Now just form the dot product of those two vectors:

(-x, -y, 1-z) . (-x, -y, -1-z) = x^2 +y^2 + (1-z)*(-1-z)

Now the truth of your claim will be embodied in that dot product being zero:

x^2 + y^2 - (1+z)(1-z) = 0

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Since R>0 and AC is perpendicular to BC, the <ACB is at right angle.

Please, find attached a simple image to show antipodal points (Two points that makes a diameter) and a point C on the sphere.

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