Let $$\triangle ABC$$ be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that $$\angle AOB = 105^\circ$$, then $$\frac{AD}{BE}$$ equals

Given that AB = AC => Angle C = Angle B (1)

AD and BE are altitudes => they make 90 degrees with the sides

Angle AOB = 105 => Angle EOD = 105 (Vertically Opposite Angles)

In quadrilateral DOEC

Angle C = 360 - 105 - 90 - 90 = 75 => Angle B = 75 (from 1)

We know that from the area of the triangle AD * BC = BE * AC

=> $$\dfrac{AD}{BE}=\dfrac{AC}{BC}=\dfrac{2R\sin B}{2R\sin A}=\dfrac{\sin\left(75\right)}{\sin\left(30\right)}=2\sin\left(75\right)$$ = 2cos(15)

[sin(x) = cos(90-x)]