Area del triangolo..

Angolo su $\small \hat{A}= \dfrac{\cancel{\pi}}{\cancel3_1}×\dfrac{\cancel{180}^{60}}{\cancel{\pi}} = 60°;$

angolo su $\small \hat{C}= \sin^{-1}\left(\dfrac{3}{5}\right) \approx{36,87°};$ $\small \;^{(1)}$

angolo su $\small \hat{B}= 180-60-36,87 = 83,13°;$

lato incognito: $\small \overline{AC}= \sqrt{20^2+(8\sqrt3)^2-2×20×8\sqrt3 ×cos(83,13)} \approx{22,928}\,u$ (teorema di Carnot);

semiperimetro $\small p= \dfrac{20+8\sqrt3+22,928}{2}\approx{28,392};$

area $\small A= \sqrt{28,392(28,392-20)(28,392-8\sqrt3)(28,392-22,928)}\approx{137,56}\,u^2$ (formula di Erone).

Note:

$\small \;^{(1)} \; \sin^{-1} = $ arcoseno.  

Th seni:

c/SIN(γ) = a/SIN(α)

fornisce una identità numerica:

8·√3/(3/5) = 20/SIN(pi/3)

40·√3/3 = 40·√3/3

Quindi:

40·√3/3 = b/SIN(β)

SIN(β) = SIN(pi - (α + γ))

SIN(β) = SIN(α + γ)

SIN(α + γ) = SIN(α)·COS(γ) + SIN(γ)·COS(α)

SIN(α) = SIN(pi/3) = √3/2

COS(α) = √(1 - (√3/2)^2) = 1/2

SIN(γ) = 3/5

COS(γ) = √(1 - (3/5)^2) = 4/5

SIN(β) = √3/2·(4/5) + 3/5·(1/2) = 2·√3/5 + 3/10

b = 40·√3·SIN(β)/3

b = 40·√3·(2·√3/5 + 3/10)/3 = 4·√3 + 16

Α = 1/2·(4·√3 + 16)·(8·√3)·SIN(pi/3)

Α = 24·√3 + 96

verifica 

Α = 1/2·(4·√3 + 16)·20·(3/5)

Α = 24·√3 + 96

Dovrebbe essere semplice