P is a point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1, F1 and F2 are the focus. If ∠ f1pf2 = 30 °, then the area of △ f1pf2 is Why let Pf1 = m, PF2 = 10-m, and then use Pythagorean theorem to calculate that there is no solution Wrong number for ∠ f1pf2 = 90 °

P is a point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1, F1 and F2 are the focus. If ∠ f1pf2 = 30 °, then the area of △ f1pf2 is Why let Pf1 = m, PF2 = 10-m, and then use Pythagorean theorem to calculate that there is no solution Wrong number for ∠ f1pf2 = 90 °

Make the vertical line of Pf1 through F2, and the vertical foot is a
Let PF2 = m and Pf1 = 10-m
PA=m/2
AF1 = 10-m-m / 2 * root 3
6 * 6 = | F1F2 | ^ 2 = | AF1 | ^ 2 + | af2 | ^ 2, can get m, can calculate slowly, very tired
Simple algorithm:
Cosine theorem 36 = | F1F2 | ^ 2 = | AF1 | ^ 2 + | af2 | ^ 2-2 × cos ∠ f1pf2 ×| AF1 | ×| af2|
Area of triangle f1pf2 = f1p × sin ∠ f1pf2 × F2P / 2