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As shown in the figure, in isosceles △ ABC, the bisectors of ∠ a = 80 °, B and C intersect at point O (1) to connect OA, and then calculate the degree of ∠ OAC; (2) to calculate: ∠ BOC
(1) Connecting Ao, ∵ in isosceles △ ABC, bisectors of ∵ B and ∵ C intersect at point O, ∵ isosceles △ ABC is symmetrical with respect to the straight line where line segment Ao is located, ∵ a = 80 °, ∵ OAC = 40 ° (2) ∵ Bo, CO bisectors ∵ ABC and ∵ ACB, respectively, ∵ OBC = 12 ∵ ABC, ∵ OCB = 12 ∵ ACB, ∵ BOC = 180 ° - (∵ OBC + ∵ OCB) = 180 °-（ 12 ‰ ABC + 12 ‰ ACB) = 180 ° - 12 (‰ ABC + ACB) = 180 ° - 12 (180 ° - a) = 90 ° + 12 ‰ A. ｜ when ‰ a = 80 °, the ‰ BOC = 180 °− 12 (‰ B + ‰ C) = 90 ° + 12 ‰ a = 130 °
It is known that in a non isosceles triangle ABC, the angle a B C and its opposite edge a B C satisfy the condition (2acosc ccosa) = a ^ 2-C ^ 2
Finding the side length of B
If a, B and C form an arithmetic sequence, and the area of △ ABC is s = root 3 / 3, calculate the perimeter of △ ABC
The wrong condition should be: 2 (acosc ccosa) = a ^ 2-C ^ 2
2(acosC-ccosA)= （2abcosC-2cbcosA）/b=（a²+b²-c²）/b--（c²+b²-a²）/b=2（a²-c²）/b
We also know that 2 (acosc ccosa) = a ^ 2-C ^ 2, so 2 (A & # 178; - C & # 178;) / b = (A & # 178; - C & # 178;), so B = 2
Because a, B and C are arithmetic sequences, B = 60 degrees
The area of △ ABC is s = acsinb / 2 = radical 3 / 3, AC = 4 / 3
A & # 178; + C & # 178; - B & # 178; = 2accosb = AC, a & # 178; + C & # 178; + 2Ac = 3aC + B & # 178; = 8, (a + C) &# 178; = 8, a + C = 2, radical 2
So the perimeter of △ ABC = a + C + B = 2 + 2, root 2
In non isosceles triangle ABC, the opposite sides of angles a, B and C are ABC and A2 = B (B + C)
(1) Verify a = 2B
(2) If a = root 3b, try to judge the shape of triangle ABC
As shown in the picture
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