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

Answer: (C)

2cos15º

Answer 2

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): $2cos\ 15°$