Same Height and Base

Label the points as above. Observe that the altitude of the isosceles triangle extends to the diameter of the circle.

Consider power of a point applied to the point $O$ . We know that $OA \times OC = OB \times OD$ .

Thus $OD = \frac{ 20 \times 20 } { 40} = 10$ . Hence, the diameter is $BD = BO+OD = 40 + 10 = 50$ .

Why go for trigonometry , when you can use pure geometry .

Consider the following. We can say that the center of circle is in the altitude as,

A line from the center of a circle perpendicular to the chord bisects the chord. So, as the chord is perpendicular and bisects the base (as it's isosceles), it's the center of the circle.

$\tan(\angle ABC) = \dfrac{40}{20} = 2 \\ \angle ABC = \tan^{-1}(2) \\ \implies \angle ABC = \angle ACB = \tan^{-1}(2) \\ \text{Using angle sum property, } \angle BAC = 180 - 2\tan^{-1}(2) \\ \implies \angle BOC = 2\angle BAC \text{, using theorem} \\ \text{Let AD be the perpendicular, } \\ \implies \angle BOD = \angle BAC = 180 - 2\tan^{-1}(2) = 2(90 - \tan^{-1}(2)) \\ \sin(\angle BOD) = \dfrac{20}{OB} = \dfrac{20}{r} \text{, r is radius since O is the center of circle and B is a point on the circumference of the circle.} \\ \sin(2(90-\tan^{-1}(2))) = \dfrac{20}{OB} \\ \sin(2\tan^{-1}(2)) = \dfrac{20}{OB} \\ \dfrac{2\times 2}{2^2 + 1} = \dfrac{20}{OB} \\ \dfrac{2}{5} = \dfrac{10}{OB} \\ \boxed{\therefore 2OB = d = 50}$