a) cot∠B = ⇔ cot 60° ⇒ x = cot 60° × 6 = 2√3
b) sin∠C = ⇔ sin∠C = ⇒ ∠C = 30°
c) ΔABC is a isosceles right triangle ⇒ AB = BC
[tex]AB^{2}+ BC^{2} =AC^{2} => 2AB^2=AC^2 \\=>AB^2=\frac{AC^2}{2} = \frac{(5\sqrt{2} )^2}{2} =25\\=>AB=x=\sqrt{25}=5 cm[/tex]
Ok done. Thank to me :>
