A triangle ABC satisfies $$\mathrm{AB} = 7, \quad \mathrm{BC} = 8, \quad \mathrm{CA} = 6.$$ We denote the center and the radius of the circumscribed circle of this triangle ABC by O and $r$, respectively. We draw two straight lines which are tangent to this circumscribed circle at the points B and C, and denote the point of intersection of these straight lines by D. We see that $$\cos\angle\mathrm{BAC} = \frac{\mathbf{A}}{\mathbf{B}}, \quad \sin\angle\mathrm{BAC} = \frac{\sqrt{\mathbf{CD}}}{\mathbf{E}},$$ $$r = \frac{\mathbf{FG}\sqrt{\mathbf{HI}}}{\mathbf{JK}}, \quad \mathrm{BD} = \mathbf{LM}.$$ Furthermore, if P is a point on the circumscribed circle, the shortest possible length of the segment DP is $\frac{\mathbf{NO}\sqrt{\mathbf{PQ}}}{\mathbf{R}}$.
A triangle ABC satisfies
$$\mathrm{AB} = 7, \quad \mathrm{BC} = 8, \quad \mathrm{CA} = 6.$$
We denote the center and the radius of the circumscribed circle of this triangle ABC by O and $r$, respectively. We draw two straight lines which are tangent to this circumscribed circle at the points B and C, and denote the point of intersection of these straight lines by D.
We see that
$$\cos\angle\mathrm{BAC} = \frac{\mathbf{A}}{\mathbf{B}}, \quad \sin\angle\mathrm{BAC} = \frac{\sqrt{\mathbf{CD}}}{\mathbf{E}},$$
$$r = \frac{\mathbf{FG}\sqrt{\mathbf{HI}}}{\mathbf{JK}}, \quad \mathrm{BD} = \mathbf{LM}.$$
Furthermore, if P is a point on the circumscribed circle, the shortest possible length of the segment DP is $\frac{\mathbf{NO}\sqrt{\mathbf{PQ}}}{\mathbf{R}}$.