Find a formula for sin A, in terms of b and c only. (N.B. it's an isosceles triangle) Solution Sin A = hc\frac{h}{c}ch. h=c2−(b2)2h = \sqrt{{c^2}-(\frac{b}{2})^2}h=c2−(2b)2 = c2−b24\sqrt{{c^2}-\frac{{b^2}}{4}}c2−4b2 = 4c2−b24\sqrt{\frac{4c^2-b^2}{4}}44c2−b2 = 4c2−b22\frac{\sqrt{4c^2-b^2}}{2}24c2−b2. So sin A = 4c2−b22c\frac{\sqrt{4c^2-b^2}}{2c}2c4c2−b2
