$\cos 30 = \frac{6}{z} \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{6}{z}$
$ \Rightarrow z = \frac{{12}}{{\sqrt 3 }} = \frac{{12}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{12\sqrt 3 }}{3} = 4\sqrt 3 $
${\left( {4\sqrt 3 } \right)^2} = {6^2} + {y^2}$
$ \Rightarrow {y^2} = {\left( {4\sqrt 3 } \right)^2} - {6^2} \Rightarrow {y^2} = 12 \Rightarrow y = \sqrt {12} = 2\sqrt 3 $
روبروی زاویه 30 درجه نصف وتر است.
$AC = 12 \Rightarrow AH = \frac{{12}}{2} = 6$
${12^2} = {6^2} + {x^2} \Rightarrow {x^2} = {12^2} - {6^2} = 108$
$x = \sqrt {108} = 6\sqrt 3 = 10/39$