It is known that: in the triangle ABC, the opposite sides of angle a, angle B and angle c are a, B, C respectively, which satisfies the square of a + the square of B + the square of C + 338 = 10A + 24B + 26c

It is known that: in the triangle ABC, the opposite sides of angle a, angle B and angle c are a, B, C respectively, which satisfies the square of a + the square of B + the square of C + 338 = 10A + 24B + 26c

The equation is reduced to a 2 - 10A + 25 + B 2 - 24B + 144 + C? - 26c + 169 = 0, so (a-5) 2 + (B-12) 2 + (C-13) 2 = 0, so a-5 = 0b-12 = 0c-13 = 0, the solution is a = 5, B = 12, C = 13. Because a  B  2 = C  2, the △ ABC is a right triangle

It is known that a, B and C are the three sides of triangle ABC, and (a-b) (a ^ 2 + B ^ 2-C ^ 2) = 0. Is triangle ABC a right triangle? Explain the reason

Because (a-b) (a ^ 2 + B ^ 2-C ^ 2) = 0
So A-B = 0 or a ^ 2 + B ^ 2-C ^ 2 = 0
So a = B, or a ^ 2 + B ^ 2 = C ^ 2
So triangle ABC is an isosceles triangle or a right triangle

If a, B and C are the three sides of △ ABC and A2 + B2 + C2 + 50 = 6A + 8b + 10C, judge the shape of the triangle

According to the known conditions, the original formula can be transformed into (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
Then the triangle is a right triangle

It is known that the three sides ABC of △ ABC satisfy a + B = 8, ab = 4, C = 2, root sign 14. Find: △ ABC is a right triangle

It can be concluded from the meaning of the title
a^2+b^2=(a+b)^2-2ab=64-8=56=c^2
So △ ABC is a right triangle with C side as the hypotenuse

If a, B and C are the three sides of △ ABC and A2 + B2 + C2 + 50 = 6A + 8b + 10C, judge the shape of the triangle

According to the known conditions, the original formula can be transformed into (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
Then the triangle is a right triangle

If a, B, C are the three sides of △ ABC, try to judge whether the value of the algebraic formula (A2 + b2-c2) 2-4a2b2 is positive or negative?

（a2+b2-c2）2-4a2b2
=（a2+b2-c2+2ab）（a2+b2-c2-2ab）
=[（a+b）2-c2][（a-b）2-c2]
=（a+b+c）（a+b-c）（a-b-c）（a-b+c），
∵ a, B, C are the three sides of the triangle ABC,
∴a+b+c＞0，a+b-c＞0，a-b-c＜0，a-b+c＞0，
If (a + B + C) (a + B-C) (a-b-c) (a-b + C) < 0, the value is negative

It is known that a, B, C are the three sides of the triangle ABC, and the shape of the triangle ABC is if a? + AB? + BC? = B? + a? B + AC

∵ A3 + AB2 + BC2 = B3 + A2B + ac2 ᙽ A3 + AB2 + bc2-b3-a2b-ac2 = 0 (a3-a2b) + (ab2-b3) - (ac2-bc2) = 0a2 (a-b) + B2 (a-b) - C2 (a-b) = 0 (a-b) (A2 + b2-c2) = 0, a = B or A2 + B2 = C2 △ ABC is a right triangle or isosceles triangle

If the lengths of the three sides of △ ABC are a, B, C, respectively, the shape of △ ABC is judged when B  2 + 2Ab = C  2 + 2Ac

∵b²+2ab=c²+2ac
∴b²-c²=2ac-2ab
∴(b-c)(b+c)=2a(c-b)
∵b+c>0,-2a

It is known that the opposite sides of angles a, B and C in the triangle ABC are a, B, C, A2 + c2-b2 = 1 / 2Ac respectively. If B = 2, find the maximum value of the triangle ABC

∵a²+c²-b²=(1/2)*ac
And the cosine theorem
cosB=(a²+c²-b²)/2ac
∴ (1/2)*ac=2ac*cosB
Then CoSb = 1 / 4
So SINB = √ 15 / 4
∵a²+c²-b²=(1/2)*ac
∴a²+c²=(1/2)*ac+b²
And a 2 + C 2 ≥ 2Ac
(if and only if a = C, obtain "=")
∴(1/2)*ac+b²≥2ac
∴ ac≤(2/3)*b²=(2/3)×2²=8/3
Area of △ ABC
S=(1/2)*ac*sinB≤(1/2)×(8/3)×(√15/4)=√15/3
Therefore, if and only if a = C, the area of △ ABC has a maximum value, and the maximum value is √ 15 / 3

It is known that a, B, C are the three side lengths of △ ABC. When B2 + 2Ab = C2 + 2Ac, try to judge which kind of triangle △ ABC belongs to and explain the reason

∵b2+2ab=c2+2ac，
∴b2+2ab+a2=c2+2ac+a2，
∴（b+a）2=（c+a）2，
∵ a, B, C are the three sides of ᙽ ABC,
/ / A, B and C are all positive numbers,
∴b+a=c+a，
∴b=c，
The triangle is an isosceles triangle