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

The Correct Option is A

To solve for \(\frac{AD}{BE}\), we must consider the properties of triangle \(ABC\) and its altitudes.

Given:

• \(\angle BAC = 45\degree\)
• \(\angle ABC = \theta\)

Since triangle \(ABC\) is a right-angled triangle and \(AD\) and \(BE\) are altitudes, we use trigonometric ratios to find their relationship. By dropping altitudes, we essentially split the triangle into smaller triangles:

• In triangle \(ABD\), angle \(\angle BAD = 45\degree\)
• In triangle \(ABE\), angle \(\angle ABE = \theta\)

Using the definition of sine in these right triangles, we have:

\(AD = AB \cdot \sin 45\degree = \frac{AB}{\sqrt{2}}\)

\(BE = AB \cdot \cos \theta\)

Therefore, the ratio is:

\[\frac{AD}{BE} = \frac{AB \cdot \frac{1}{\sqrt{2}}}{AB \cdot \cos \theta} = \frac{1}{\sqrt{2} \cdot \cos \theta}\]

Multiplying the numerator and denominator by \(\sqrt{2}\) gives:

\[\frac{\sqrt{2}}{2 \cdot \cos \theta} = \sqrt{2} \cdot \sin \theta\]

Thus, the correct answer is:

\(\sqrt{2}\sin\theta\)

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): \(\sqrt2sin\theta\)