1, the diameter CD of the circle passing through point c is connected with BD, and by using the property that the same arc in the same circle corresponds to the same circumferential angle, ∠D=∠A, ∠ DBC = 90, then BC: CD = 3: 4, and the solution is CD= 16, so the radius R.

Generally speaking, in the same big question, the second question often uses the conclusion of the first question, but this is a guessing question. Let me first talk about my own ideas for your reference.

(1), the first question By making auxiliary lines, we get the conclusion that sin∠A=a/2R, where A = BC and r is the radius of a circle;

② In the same way, it is concluded that sin∠B=b/2R and sin∠C=c/2R, where b=AC and c = ab.

(3) If the above conclusion is too simple, it is mainly to realize that all three angles are acute angles, which is equivalent to sine. We also know that ∠A+∠B+∠C=∏, and further say that there is SIN∠C = SIN(∠A+∞).

④ One conclusion I put forward is: Please prove: c=a√ 1-(b/2R)? + a√ 1-(b/2R)？ ;

⑤ The proof is very simple, so I won't explain it in detail here, but for your junior high school students, I want to add a sum-difference product formula: SIN (∠ A+∠ B) = SIN ∠ ACOS ∠ B+COS ∠ ASIN ∠ B.