In △ ABC, ab = 5, BC = 3, when ∠ ABC = what, the area of △ ABC is the largest

In △ ABC, ab = 5, BC = 3, when ∠ ABC = what, the area of △ ABC is the largest

S △ ABC = 1 / 2Sin ∠ ABC * AB * BC, so when sin ∠ ABC reaches the maximum, the area is the largest, that is, when sin ∠ ABC = 1, the area is the largest when sin ∠ ABC = 90

In △ ABC, ab = 2, AC = 3, BC = 4, then the area of △ ABC is

cosA=(2^2+3^2-4^2)/2*2*3=-1/4
Sina = root 15 / 4
S = 1 / 2 * 2 * 3 * root 15 / 4 = 3 (root 15 / 4)

S is the area of △ ABC, AB ^ 2 + BC ^ 2 + AC ^ 2 > = 4 √ 3 s
How to use coordinate method
It's the root

From the cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bccosa
From the area formula: S = (1 / 2) bcsina
Difference method:
a^2+b^2+c^2-4√3S
=b^2+c^2-2bccosA+b^2+c^2-4√3*(1/2)bcsinA
=2b^2+2c^2-2bccosA-2√3bcsinA
=2b^2+2c^2-4bc[(1/2)cosA+(√3/2)sinA]
=2b^2+2c^2-4bc+4bc-4bccos(60-A)
=2(b-c)^2+4bc[1-cos(60-A)]
-120 < 60-A < 60
-1/2 < cos(60-A) ≤ 1
0 ≤ 1-cos(60-A) < 3/2
therefore
a^2+b^2+c^2-4√3S = 2(b-c)^2+4bc[1-cos(60-A)] ≥0
When B = C and a = 60, i.e. equilateral triangle, the equal sign holds

In △ ABC, ab = AC, BC = 8, if the area of △ ABC is 8, then AB = AC, BC = 8

The height of BC is 2, AB * * 2 = 4 * * 2 + 2 * * 2, so AB = √ 20