Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 244: 81

$AD = \sqrt{6}+\sqrt{2}$

We can use the Pythagorean theorem to find the length of $AD$: $(AD)^2 = (AC)^2+(DC)^2$ $AD = \sqrt{(AC)^2+(DC)^2}$ $AD = \sqrt{(1)^2+(2+\sqrt{3})^2}$ $AD = \sqrt{1+(4+4\sqrt{3}+3)}$ $AD = \sqrt{8+4\sqrt{3}}$ $AD = \sqrt{6+2\sqrt{12}+2}$ $AD = \sqrt{(\sqrt{6}+\sqrt{2})~(\sqrt{6}+\sqrt{2})}$ $AD = \sqrt{6}+\sqrt{2}$