Let 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 $\angle AOB=105$, then $\frac{AD}{BE}$ equals

The correct option is 2cos15.

Let us solve the problem step-by-step:

Step 1: Understand the geometric properties of the triangle
We are given that $ABC$ is an isosceles triangle with $AB=AC$.
Since $AD$ is the altitude from the vertex $A$ to the base $BC$ in an isosceles triangle, it also acts as the angle bisector of $\angle A$.
Let $\angle A=2\theta$.
Therefore, the angle bisected by the altitude is:
$\angle BAD=\angle CAD=\theta$

Step 2: Relate the angles in the triangle to find $\theta$
Now consider the altitude $BE$ from vertex $B$ to side $AC$. Since $BE\perp AC$, the triangle $AEB$ is a right-angled triangle at $E$.
Thus, in $△AEB$, we have:
$\angle ABE=90°-\angle A=90°-2\theta$

The intersection of altitudes $AD$ and $BE$ is $O$. In $△AOB$, the sum of the interior angles is $180°$:
$\angle BAO+\angle AOB+\angle ABO=180°$

Substituting the known angles in terms of $\theta$ and the given value $\angle AOB=105°$:
$\theta +105°+(90°-2\theta )=180°$
$195°-\theta =180°$
$\theta =15°$

Thus, the vertex angle is $\angle A=2\theta =30°$.

Step 3: Express the altitudes $AD$ and $BE$ in terms of the side length
Let the length of the equal sides be $AB=AC=c$.
In right-angled triangle $ABD$:
$AD=ABcos(\angle BAD)=ccos\theta =ccos15°$

In right-angled triangle $ABE$:
$BE=ABsin(\angle A)=csin\left(2\theta \right)=csin30°$

Step 4: Find the ratio $\frac{AD}{BE}$
Now, let's divide $AD$ by $BE$:
$\frac{AD}{BE}=\frac{ccos15°}{csin30°}=\frac{cos15°}{sin30°}$

Since $sin30°=\frac{1}{2}$, we have:
$\frac{AD}{BE}=\frac{cos15°}{1/2}=2cos15°$