Question

In triangle $ABC$, altitudes $AD$ and $BE$ are drawn to the corresponding bases. If $∠BAC=45\degree$ and $∠ABC=θ$, then $\frac {AD}{BE}$ equals

• A. $\sqrt2sin\theta$
• B. $\sqrt2cos\theta$
• C. $\frac{(sinθ+cosθ)}{\sqrt2}$
• D. 1

Answer 1

Answer: (A)

$\sqrt2sin\theta$

Answer 2

To find the value of $\frac {AD}{BE}$, we'll use the given information about the angles in $△ ABC$ and the properties of altitudes.
Let's start by labeling the triangle and its angles:
$∠BAC = 45°$
$∠ABC = θ$
Now, let's draw altitudes AD and BE:

Angle $\angle BAE = 45°$ degrees is stated.
This suggests $AE = BE$.
Suppose $AE = BE = x$.
It is written as $\angle{ABC}=\theta$ in the right-angled $△ ABD$.
$sin\theta=\frac{AD}{AB}$

$sin\theta=\frac{AD}{x\sqrt{2}}$

$\sqrt2sin\theta=\frac{AD}{BE}$

$\frac{AD}{BE}=\sqrt2sin\theta$

So, the correct option is (A): $\sqrt2sin\theta$