cat 2023 Question 17

Question

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 $\angle \mathrm{AOB}=105^{\circ}\angle \mathrm{AOB}=105^{\circ}$, then $\frac{A D}{B E}$ equals

Accepted Answer

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

=> $\frac{AD}{BE} = \frac{AC}{BC} = \frac{2R Sin(B)}{2R Sin(A)} = \frac{Sin(75)}{Sin(30)} = 2 Sin(75) = 2 Cos(15)$

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

Question 17.

Question Explanation