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

$$\frac{{\pi {{\text{R}}^2}{\text{E}}}}{{{{90}^ \circ }}}$$

$$\frac{{\pi {{\text{R}}^2}{\text{E}}}}{{{{180}^ \circ }}}$$

$$\frac{{\pi {{\text{R}}^2}{\text{E}}}}{{{{270}^ \circ }}}$$

$$\frac{{\pi {{\text{R}}^2}{\text{E}}}}{{{{360}^ \circ }}}$$

If E is the spherical excess and R the radius of the earth, the surface area of the triangle, is Correct Answer $$\frac{{\pi {{\text{R}}^2}{\text{E}}}}{{{{180}^ \circ }}}$$

Bissoy MCQ Team

Civil Engineering
Surveying

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

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