CAT 2023 Slot 3 QA Question 18: Let ΔABC 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^

Question

Let ΔABC 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^

Accepted Answer

Since AD and BE are altitudes, O is the orthocentre of ΔABC.

In any triangle, $\angle AOB = 180^\circ - \angle C$ when O is the orthocentre.

$105^\circ = 180^\circ - \angle C$
$\angle C = 75^\circ$

Since $AB = AC$, the base angles are equal.

$\angle B = \angle C = 75^\circ$
$\angle A = 30^\circ$

Let ΔABC 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 ∠AOB=105∘\angle AOB = 105^\circ∠AOB=105∘, then ADBE\dfrac{AD}{BE}BEAD​ equals

Let ΔABC 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 $\dfrac{AD}{BE}$ equals