Question:medium

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 \(∠AOB =105\degree\) , then \(\frac{AD}{BE}\) equals

Updated On: Jan 13, 2026
  • sin15º
  • cos15º
  • 2cos15º
  • 2sin15º
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The Correct Option is C

Solution and Explanation

Let ΔABC be an isosceles triangle such that AB and AC are of equal length

Given, \( AB = AC \) implies \( \angle C = \angle B \) (Equation 1). \( AD \) and \( BE \) are altitudes, meaning they are perpendicular to their respective sides. \( \angle AOB = \angle EOD = 105^\circ \) due to vertically opposite angles. In quadrilateral \( DOEC \), the sum of angles is \( 360^\circ \). Therefore, \( \angle C = 360^\circ - 105^\circ - 90^\circ - 90^\circ = 75^\circ \). From Equation 1, \( \angle B = 75^\circ \). The area of a triangle can be expressed as \( AD \cdot BC = BE \cdot AC \). Rearranging this gives \( \frac{AD}{BE} = \frac{AC}{BC} \). Using the sine rule, \( \frac{AC}{BC} = \frac{2R \sin B}{2R \sin A} \). Thus, \( \frac{AD}{BE} = \frac{\sin 75^\circ}{\sin 30^\circ} \). This simplifies to \( \frac{AD}{BE} = 2 \sin 75^\circ \), which is equivalent to \( \frac{AD}{BE} = 2 \cos 15^\circ \).

Correct option: (C) \( 2 \cos 15^\circ \)

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