Given:
A figure of a triangle, AD = 6 and DB = 2.
To find:
The length of AC.
Solution:
In a right angle triangle,
[tex]\cos \theta =\dfrac{Base}{Hypotenuse}[/tex]
In triangle ACD,
[tex]\cos A =\dfrac{AD}{AC}[/tex]
[tex]\cos 30^\circ =\dfrac{6}{AC}[/tex]
[tex]\dfrac{\sqrt{3}}{2}=\dfrac{6}{AC}[/tex]
[tex]\sqrt{3}\times AC=6\times 2[/tex]
[tex]AC=\dfrac{12}{\sqrt{3}}[/tex]
[tex]AC=\dfrac{12}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}[/tex]
[tex]AC=\dfrac{12\sqrt{3}}{3}[/tex]
[tex]AC=4\sqrt{3}[/tex]
Therefore, the length of AC is [tex]4\sqrt{3}[/tex] units.