The figure to the right is a net for the tetrahedron OABC. This tetrahedron satisfies
$$\begin{gathered} \mathrm { BC } = 10 , \quad \mathrm { AC } = 8 , \quad \sin \angle \mathrm { ACB } = \frac { 3 } { 4 } , \\ \mathrm { OA } = 4 , \quad \triangle \mathrm { ABC } \equiv \triangle \mathrm { OBC } . \end{gathered}$$
(1) The area of the triangle ABC is $\mathbf { A B }$.
(2) Let AH denote the perpendicular line drawn from point A to side BC. The length of AH is $\mathbf { C }$.
(3) Let $\theta$ denote the angle formed by the plane ABC and the plane OBC. Then we have
$$\cos \theta = \frac { \mathbf { D } } { \mathbf { E } } , \quad \sin \theta = \frac { \mathbf { F } \sqrt { \mathbf { G } } } { \mathbf { H } } .$$
(4) The volume of the tetrahedron OABC is $\frac { \mathbf { IJ } \sqrt { \mathbf { K } } } { \mathbf { L } }$.