It so happens that the triangle whose sides are 4, 5, and 6 has a special property: Its largest angle is twice its smallest angle. Make calculations that support this assertion.

∠

*A*is the largest; ∠*C*is the smallest.Solution 1. We can use the law of cosines:

6

cos

∠

^{2}= 4^{2}+ 5^{2}– 2 · 4 · 5 cos*A*cos

*A*= (6^{2}– 4^{2}– 5^{2}) / (-2 · 4 · 5) = 0.125∠

*A*= arccos (0.125) ≈ 82.82° 4^{2} = 5^{2} + 6^{2} – 2 · 5 · 6 cos *C*

cos *C* = (4^{2} – 5^{2} – 6^{2}) / (-2 · 5 · 6) = 0.75

∠*C* = arccos (0.75) ≈ 41.41°

Solution 2. We can use area, via Heron’s formula:

Area = √

Area = √(15/2)(3/2)(5/2)(7/2)

A = (15 √7) / 4

*s*(*s – a*)(*s – b*)(*s – c*) where*s*is the semiperimeter (= 15/2)Area = √(15/2)(3/2)(5/2)(7/2)

A = (15 √7) / 4

Now we can find *h*: Area = *bh*/2 = 5*h*/2 = 15 √7 / 4 → *h* = 3 √7 / 2

And now, sin *A* = *h*/4 = 3 √7 / 8 → ∠*A* ≈ 82.82°

And sin *C* = *h*/6 = 3 √7 / 12 → ∠*C* ≈ 41.41°.

These solutions are a good advertisement for the law of cosines, don’t you think?