Can you measure the angle?

We know that $\angle BOC = 90^{\circ} - \frac{\angle A}{2}$ .

Applying Law's of cosine in $\triangle ABC$ , We have $BC^2 = AC^2 + AB^2 - 2 \times AC \times AB \times \cos \angle A$ .

Substituting the values we get $\cos \angle BAC = \frac{1}{2}$ .

Hence $\angle A = 60^{\circ}$ .

Therefore, $\angle BOC = 90^{\circ} - \frac{60^{\circ}}{2}= 90^{\circ} - 30^{\circ} = 60^{\circ}$ .

Therefore, $\angle EDC = \frac{1}{2} \times 60^{\circ} = \boxed{30^{\circ}}$

By Law of Cosines on $\triangle ABC$ , $\left(\dfrac{sqrt{2}}{\sqrt{3}-1}\right)^2=(\sqrt{2})^2+(\sqrt{3})^2-2\cdot \sqrt{2}\cdot \sqrt{3}\cos B$

Simplifying ths expression, we obtain that $2+\sqrt{3}=5-2\sqrt{6}\cos B$

Isolating $\cos B$ shows that $\cos B = \dfrac{3-\sqrt{3}}{2\sqrt{6}}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$ so $B=75^{\circ}$ .

By Law of Sines, $\dfrac{\dfrac{\sqrt{2}}{\sqrt{3}-1}}{\sin 75^{\circ}}=\dfrac{\sqrt{2}}{\sin C}$ Solving for $\sin C$ , we obtain $\sin C=\dfrac{\sqrt{2}}{2}$ so $C=45^{\circ}$ . This also means $A=60^{\circ}$ .

Note that $BO$ is the angle bisector of $\angle CBJ$ , and $CO$ is the angle bisector of $\angle BCK$ , from the definition of the excircle. Thus, $\angle BCO=\dfrac{135}{2}$ and $\angle BCO=\dfrac{105}{2}$ .

Thus, $\angle BOC=60^{\circ}$ , and $\angle EDF = \dfrac{60}{2}=\boxed{30^{\circ}}$ .

Using Heron's formula we get the area of triangle 1.183012702,Now 1/2 * BC * AC sin C=1.183..... solving it we get C=45,and the same way B=75,So JBC=105 that makes OBC=105/2=52.5[when an exterior circle is drawn the exterior angle is bisected] ,similarly OCB=67.5,then BOC=60,and inscribed is 1/2 of central angle so EDF=60/2=30

$Let\quad ,\quad in\quad triangle\quad ABC\quad ,\quad BC\quad \quad =\quad a\quad =\quad \sqrt { 3 } \quad ,\quad AC\quad \quad =\quad b\quad =\quad \frac { \sqrt { 2 } }{ \sqrt { 3 } -\quad 1 } \quad ,\quad \quad \quad \& \quad \quad AB\quad =\quad c\quad =\quad \sqrt { 2 } \\ so,\quad Cos\angle ABC\quad =\quad \frac { { c }^{ 2 }\quad +\quad { a }^{ 2 }\quad -\quad { b }^{ 2 } }{ 2ca } \quad =\quad \frac { ({ \sqrt { 2 } ) }^{ 2 }\quad +\quad { (\sqrt { 3 } ) }^{ 2 }\quad \quad -\quad { (\frac { \sqrt { 2 } }{ \sqrt { 3 } -\quad 1 } ) }^{ 2 } }{ 2\quad \times \quad \sqrt { 2 } \quad \times \quad \sqrt { 3 } } \quad =\quad \frac { \sqrt { 6 } -\quad \sqrt { 2 } }{ 4 } \quad \\ so,\quad \angle ABC\quad =\quad \quad \cos ^{ -1 }{ (\quad \frac { \sqrt { 6 } -\quad \sqrt { 2 } }{ 4 } ) } =\quad 75\quad \quad \quad \& \quad \angle JBC\quad =\quad (180\quad -\quad 75)\quad =\quad 105\quad \\ Cos\angle ACB\quad =\quad \frac { { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad -\quad { c }^{ 2 } }{ 2ab } \quad =\quad \frac { ({ \sqrt { 3 } ) }^{ 2 }\quad +\quad { (\frac { \sqrt { 2 } }{ \sqrt { 3 } -\quad 1 } ) }^{ 2 }\quad -\quad { (\sqrt { 2 } ) }^{ 2 } }{ 2\quad \times \quad \sqrt { 3 } \quad \times \quad (\frac { \sqrt { 2 } }{ \sqrt { 3 } -\quad 1 } ) } \quad =\quad \frac { 1 }{ \sqrt { 2 } } \\ so,\quad \angle ACB\quad =\quad \cos ^{ -1 }{ (\frac { 1 }{ \sqrt { 2 } } ) } \quad =\quad 45\quad \quad \quad \& \quad \angle KCB\quad =\quad (180\quad -\quad 45)\quad =\quad 135\quad \\ As\quad \quad circle\quad DIGH\quad is\quad an\quad exterior\quad circle\quad \quad ,\quad \quad so\quad \quad BO\quad \& \quad CO\quad are\quad the\quad bisectors\quad of\quad \angle JBC\quad \& \quad \angle KCB\quad \\ so,\quad \quad \angle OBC\quad =\quad \frac { \angle JBC }{ 2 } \quad =\quad \frac { 105 }{ 2 } \quad =\quad 52.5\quad \quad \quad \& \quad \angle OCB\quad =\quad \frac { \angle KCB }{ 2 } \quad =\quad \frac { 135 }{ 2 } \quad =\quad 67.5\\ In\quad triangle\quad OBC\quad \quad ,\quad \quad \angle BOC\quad =\quad 180\quad -\quad \angle OBC\quad -\quad \angle OCB\quad =\quad 180\quad -\quad 52.5\quad -\quad 67.5\quad =\quad 60\\ so,\quad \angle EOF\quad =\quad \angle BOC\quad =\quad 60\\ as\quad we\quad know,\quad an\quad inscribed\quad angle\quad is\quad half\quad of\quad the\quad central\quad angle\quad from\quad the\quad same\quad arc\quad (intercepted\quad arc)\quad \\ so,\quad \angle EDF\quad =\quad \frac { \angle EOF }{ 2 } \quad =\quad 30$