The equation for the area of a spherical triangle can be proven by finding four areas.
- Area of a
- Areas of three lunes that create the spherical triangle
- Surface area of the sphere in terms of the areas of the lunes
- Area of the spherical triangle
Area of a Spherical Lune
Let
L(x) denote the area of a spherical lune where
x is the angle between two .
The is
4πr2 and its measures
360∘. Because the area of the spherical lune is to the angle, the following can be written.
xL(x)=360∘4πr2
The equation can now be solved for
L(x) to find the area of the spherical lune.
xL(x)=360∘4πr2
xL(x)=90∘πr2
L(x)=90∘πr2⋅x
L(x)=90∘xπr2
Area of the Spherical Lunes
Consider a two-dimensional view from the top of the pair of spherical lunes. Since the are by the , a pair of opposite spherical lunes have the same area.
Each pair of great circles intersects at two opposite . Consequently, the intersection of three great circles will create two spherical triangles that are opposite to each other. Based on the diagram,
x, y, and
z are the measures of the interior angles of the spherical triangles.
Similar to
L(x), areas
L(y) and
L(z) can be found.
L(x)=90∘xπr2L(y)=90∘yπr2L(z)=90∘zπr2
Surface Area of the Sphere
To find the area
A of the spherical triangle, the surface area of the sphere will be written in terms of the area of the spherical lunes and the area of the spherical triangle. It can be noted that all three pairs of lunes cover the entire sphere.
Now, add the surface areas of the spherical lunes. Since the lunes in a pair share the same surface area, each area will be doubled.
2L(x)+2L(y)+2L(z)
Each pair of lunes intersects at the triangle and the opposite triangle, so the sum contains the area of the triangle six times. Consequently, four areas
A of the triangle must be subtracted from the sum so that the area of each triangle is calculated only once. In this way, the surface area of the sphere,
4πr2, will be obtained.
2L(x)+2L(y)+2L(z)−4A=4πr2
Area of the Spherical Triangle
Finally, the expressions for the areas of the spherical lunes will be substituted into the equation to solve for
A.
2L(x)+2L(y)+2L(z)−4A=4πr2
2(90∘xπr2)+2(90∘yπr2)+2(90∘zπr2)−4A=4πr2
45∘xπr2+45∘yπr2+45∘zπr2−4A=4πr2
45∘xπr2+45∘yπr2+45∘zπr2=4A+4πr2
45∘xπr2+45∘yπr2+45∘zπr2=4(A+πr2)
180∘xπr2+180∘yπr2+180∘zπr2=A+πr2
180∘xπr2+180∘yπr2+180∘zπr2−πr2=A
A=180∘xπr2+180∘yπr2+180∘zπr2−πr2
A=(180∘x+180∘y+180∘z−1)πr2