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## Homework Statement

Suppose [itex]T[/itex] is an equilateral triangle on the sphere of radius [itex]R = 1[/itex]. Let [itex] \alpha [/itex] denote the angle at any of the three vertices’s of the triangle. (Recall that [itex] 3\alpha > n[/itex].) Use the result of the last problem on the previous homework and the inclusion - exclusion principle (together with an orange and a knife) to compute the area of [itex]T[/itex] .

## Homework Equations

The result to the last problem on the previous homework is [itex]A = \alpha2R^2[/itex]

## The Attempt at a Solution

I assumed that all angle on the equilateral triangle where 90 degrees or [itex]\frac{\pi}{2}[/itex]; therefore making the volume equal to 1/8 that of the whole sphere

So I did

[itex]A = \alpha2R^2[/itex] where [itex]A[/itex] is the area of [itex]T[/itex]

[itex]A = \frac{\pi}{2}2R^2[/itex]

[itex]A = \pi*R^2[/itex] That would be the area of 1/4 of the sphere overall, but because I am taking the area of an equilateral triangle, I took half of that to get

[itex]A = \frac{\pi}{2}R^2[/itex]

[itex]A = \frac{\pi}{2}*1[/itex]

[itex]A = \frac{\pi}{2}[/itex]

Would that be correct? I just kind of picked 90 degrees or [itex]\frac{\pi}{2}[/itex] for [itex]\alpha[/itex], but I assume it could be anything between 60 and up to 90 degrees which would change my answer. How do I know which angle to pick?

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