SOLUTION TO LAST MONTH'S QUIZ
1. 6
2. 8
3. 8.727 cm^{2}
Notes.
3. (Spherical Trigonometry teaches us a simple relationship between the area of a spherical triangle and the excess of its anglesum over 180^{o}. But the following argument solves the problem from first principles).
If two great circles intersect at an angle q^{o}, it is easily seen that each of the two enclosed "lunes" has an area equal to q/360 of the total surface of the sphere. So if a represents the area of one of the green regions as a fraction of the whole surface, b the area of one red region, c the area of one yellow region, and x the area of triangle ABC (and of its image, A'B'C'), we have:
a + x = 45/360 = 1/8
b + x = 90/360 = 1/4
c + x = 50/360 = 5/36
Also 2a + 2b + 2c + 2x = 1
Solving, x = 1/144
But total surface area = 4 * PI * 10^{2}
So Area ABC = 4 * PI * 10^{2} / 144 = 8.7266 cm^{2} 

