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

• A. sin15º
• B. cos15º
• C. 2cos15º
• D. 2sin15º

Answer 1

The Correct Option is C

Given, \( AB = AC \) 
\( \Rightarrow \angle C = \angle B \) …….. (1)
\( AD \) and \( BE \) are altitudes \( \Rightarrow \) they make 90° with the sides.
\( \angle AOB = \angle EOD = 105^\circ \) (Vertically Opposite Angles)
In quadrilateral \( DOEC \):
\( \angle C = 360^\circ - 105^\circ - 90^\circ - 90^\circ = 75^\circ \)
From equation (1):
\( \Rightarrow \angle B = 75^\circ \)

Area of triangle:
\( AD \cdot BC = BE \cdot AC \)
\( \Rightarrow \frac{AD}{BE} = \frac{AC}{BC} \)
\( \Rightarrow \frac{AD}{BE} = \frac{2R \sin B}{2R \sin A} \)
\( \Rightarrow \frac{AD}{BE} = \frac{\sin 75^\circ}{\sin 30^\circ} \)
\( \Rightarrow \frac{AD}{BE} = 2 \sin 75^\circ \)
\( \Rightarrow \frac{AD}{BE} = 2 \cos 15^\circ \)

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