two altitudes of an isosceles triangle are equal to 20 cm and 30 cm. determine the possible measures of the base angles of the triangle.

D is the mid-point of AC, AB = BC = a and CE is the perpendicular from C to AB and is of length h.

$$\displaystyle \frac{h}{a}=\sin(\angle EBC) = \sin(180 - 2\angle BAC)=\sin(2\angle BAC)\\ \\ =2\sin(\angle BAC)\cos(\angle BAC)\\ \\=2\times \frac{30}{a}\times \frac{\sqrt{a^{2}-30^{2}}}{a}$$

If h is to equal 20, then, multiplying by a squared and cancelling,

$$\displaystyle a=3\sqrt{a^{2}-900},$$

from which, (square both sides),

$$a=\sqrt{8100/8}\approx 31.8198.$$

The base angle can then be calculated. Repeat the calculation for BD = 20, and h = 30.

Thanks for doing this one Bertie, it had me totally stumped    

Nice one, Bertie......!!!!!