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

Updated On: Jul 26, 2025
  • \(\sqrt2sin\theta\)
  • \(\sqrt2cos\theta\)
  • \(\frac{(sinθ+cosθ)}{\sqrt2}\)
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The Correct Option is A

Approach Solution - 1

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:
\(\sqrt{2}\sin\theta\)
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Approach Solution -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:
 In triangle ABC,altitudes AD and BE are drawn to the corresponding bases.
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): \(\sqrt2sin\theta\)

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