Let the two numbers be represented by x and y. The problem statement gives rise to two sets of equations.
x - y = 0.6
y/x = 0.6 . . . . . . . assuming x is the larger of the two numbers
or
x/y = 0.6 . . . . . . . assuming y has the larger magnitude
The solution of the first pair of equations is
(x, y) = (1.5, 0.9)
The solution of the first and last equations is
(x, y) = (-0.9, -1.5)
The pairs of numbers could be {0.9, 1.5} or {-1.5, -0.9}.
1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :
i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.
ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:
a^{2}=b ^{2}+c ^{2}-2bc(cosA)
2.
20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA) 400=81+169-234(cosA) 150=-234(cosA) cosA=150/-234= -0.641
3. m(A) = Arccos(-0.641)≈130°,
4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc
A would be the answer John hit the baseball it went over the fence.
