If E is the spherical excess and R the radius of the earth, the surface area of the triangle, is

- 1
  
  $\frac{{πR}^{2} E}{90^{0}}$
- 2
  
  $\frac{{πR}^{2} E}{180^{0}}$
- 3
  
  $\frac{{πR}^{2} E}{270^{0}}$
- 4
  
  $\frac{{πR}^{2} E}{360^{0}}$

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