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
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: Aug 17, 2024
- sin15º
- cos15º
- 2cos15º
- 2sin15º
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The Correct Option is C
Solution and Explanation
Given, \(AB = AC\)
\(⇒ ∠C = ∠B \) …….. (1)
\(AD\) and \(BE\) are altitudes \(⇒\) they make 90° with the sides.
\(∠AOB = EOD = 105°\) (Vertically Opposite Angles)
In quadrilateral \(DOEC\)
\(∠C = 360°-105°-90°-90°\)
\(∠C= 75°\)
From eq (1),
\(⇒ ∠B = 75°\)
Area of the triangle
\(AD.BC = BE .AC\)
\(\frac {AD}{BE}=\frac {AC}{BC}\)
\(\frac {AD}{BE} =\frac {2R sin \ B}{2R sin \ A}\)
\(\frac {AD}{BE}=\frac {sin 75°}{sin 30°}\)
\(\frac {AD}{BE}=2sin\ 75°\)
\(\frac {AD}{BE}=2cos\ 15°\)
So, the correct option is (C): \(2cos\ 15°\)
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