A geometry problem about triangles

$Let\ \angle HCA=\alpha.\ \ \ \therefore\ \angle BCK=180-60-\alpha=120-\alpha.\\ So\ \dfrac {Sin(120-\alpha)}{Sin\alpha}=Sin120*Cot\alpha-Cos120=\color{#3D99F6}{\frac1 2 *(\sqrt3*Cot\alpha+1)}.\ \ \ \ Cos120=-\frac 1 2.\\ In \ \Delta s\ AHC\ and\ BCK,\\ AC=\dfrac{AH}{Sin\alpha}=\dfrac 3 {Sin\alpha},\ \ \ and\ \ \ BC=\dfrac{BK}{Sin(120-\alpha)}=\dfrac 4 {Sin(120-\alpha)}, \\ AB=AC=BC,\\ \ \implies\ \dfrac 4 3=\dfrac{Sin(120-\alpha)}{ Sin\alpha}=\frac1 2 *(\sqrt3*Cot\alpha+1).\\ \therefore\ Cot\alpha=\dfrac{\dfrac 8 3 -1}{\sqrt3}=\dfrac 5 {3\sqrt3}.\\ \therefore\ \dfrac 1 {Sin\alpha^2}=( \dfrac 5 {3\sqrt3} )^2+1\\ AB=AC=\dfrac 3 {Sin\alpha}=3\sqrt{\dfrac {25+27}{3\sqrt3}}=\Large\ \ \ \color{#D61F06}{4.16.}$

(1) Draw a line through A parallel to HK, constructing a right triangle with sides a + b, 1, and S = AB. Then ((a + b)^2 + 1 = S^2, or a^2 + 2ab + b^2 + 1 = S^2. (2) From the right triangles, we have: a^2 + 9 = S^2 and b^2 + 16 = S^2. (3) Solving for 2ab from (1), and substituting for a^2 and b^2 from (2), 2ab = S^2 -1 - (S^2 -9) - (S^2 - 16), or 2ab = 24 - S^2. (4) Squaring, 4a^2b^2 = 576 - 48S^2 + S^4 = 4(S^2 - 9)(S^2 - 16) = 4S^4 - 100S^2 + 576. Equating these expressions, 3S^2 = 52, and S = 4.163. Ed Gray