The two equations of X (M + 5) * 2 - (2m-5) + M-8 = 0, (1) find the value of M; (2) find the length of three sides of triangle ABC Sorry, the last question is wrong, now add. Thank you, you master! It is known that ⊿ ABC, a, B and C are the opposite sides of ⊿ A and ⊿ B ⊿ C (a > b) of triangle ABC, and a ^ 2 + B ^ 2 = C ^ 2. Sina and AINb are two of the equations (M + 5) x ^ 2 - (2m-5) x + M-8 = 0 of X respectively, (1) find the value of M; (2) find the length of three sides of ⊿ ABC

The two equations of X (M + 5) * 2 - (2m-5) + M-8 = 0, (1) find the value of M; (2) find the length of three sides of triangle ABC Sorry, the last question is wrong, now add. Thank you, you master! It is known that ⊿ ABC, a, B and C are the opposite sides of ⊿ A and ⊿ B ⊿ C (a > b) of triangle ABC, and a ^ 2 + B ^ 2 = C ^ 2. Sina and AINb are two of the equations (M + 5) x ^ 2 - (2m-5) x + M-8 = 0 of X respectively, (1) find the value of M; (2) find the length of three sides of ⊿ ABC

From a ^ 2 + B ^ 2 = C ^ 2, we know that ⊿ ABC is a right triangle, and ∠ C is a right angle Sina. SINB are two Sina + SINB = (2m-5) / (M + 5) Sina * SINB = (M-8) / (M + 5) and Sina ^ 2 + SINB ^ 2 = 1 of the equation (M + 5) x ^ 2 - (2m-5) x + M-8 = 0 of X respectively. We can get m ^ 2-24m + 80 = 0m = 20 or M = 4. Through the test, M = 20 can be obtained because a > b is (x-3 / 5)