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The eighth grade needs details to solve a mathematical geometry application problem (about isosceles triangle)
( 1)∠A= 1 10? ∠ b = 20, ∠ c = 50. When △PAC becomes an isosceles triangle, there are two situations:

When PC=AC, the vertex angle ∠ ACP = 50, as shown in the right figure; When AP=AC, the vertex angle ∠ PAC = 80.

Or ∠ p ˊ AC = 1 10, as shown in the left figure; When PA=PC, the vertex angle ∠ APC = 80.

To sum up: when c is the vertex of the vertex angle, the vertex angle is 50; When a is the vertex of the vertex angle, the vertex angle is 1 10 or 80.

When p is the vertex of the vertex angle, the vertex angle is 80.

(2) In △ ABC? AB=AC= 17? BC= 16? O is the midpoint of BC, which is obtained by Pythagorean theorem.

OA? = 17? -8? =225, OA= 15, let AB equation be y=kx+b, and let A (0, 15) and B (-8, 0).

Substituting the results b= 15 and k= 15/8, then the analytical formula of straight line AB is y= 15x/8+ 15.

(3) Set the X roof of the tent, which is 3x+7≤5(x- 1)+3, 2x≥9.

Because the top of tent X is an integer, the number of tents x=5 and the number of people 5*3+7=22.