Question

Column A: \(\frac{BD}{AB}\)
Column B: \(\frac{DC}{BC}\)

• A. Quantity A is greater
• B. Quantity B is greater
• C. The two quantities are equal
• D. The relationship cannot be determined from the information given

Hint:

For geometry problems where dimensions or angles are not given, test extreme or special cases (e.g., isosceles, right, or very skewed triangles) to see if the relationship between the quantities changes.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem asks us to compare two ratios of side lengths in a triangle that has been divided into two right-angled triangles by an altitude. We can use trigonometric ratios or test different cases to determine the relationship.
Step 2: Key Formula or Approach:
In a right-angled triangle, the sine and cosine of an angle are defined as:
\(\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
Let's analyze the ratios in the two right-angled triangles, \(\triangle ADB\) and \(\triangle CDB\).
In right \(\triangle ADB\), with respect to angle A:
Column A: \(\frac{BD}{AB} = \frac{\text{Opposite side to A}}{\text{Hypotenuse}} = \sin(A)\).
In right \(\triangle CDB\), with respect to angle C:
Column B: \(\frac{DC}{BC} = \frac{\text{Adjacent side to C}}{\text{Hypotenuse}} = \cos(C)\).
So the problem is to compare \(\sin(A)\) and \(\cos(C)\).
Step 3: Detailed Explanation:
The problem provides no information about the angles A and C, or the side lengths of \(\triangle ABC\). Therefore, the relationship between \(\sin(A)\) and \(\cos(C)\) can vary. Let's test a few cases with specific triangles.
Case 1: Let \(\triangle ABC\) be an isosceles right triangle with \(\angle B = 90^{\circ}\) and \(AB = BC\).
In this case, the altitude BD from B to AC makes \(\angle A = \angle C = 45^{\circ}\).
\(\sin(A) = \sin(45^{\circ}) = \frac{\sqrt{2}}{2}\).
\(\cos(C) = \cos(45^{\circ}) = \frac{\sqrt{2}}{2}\).
In this scenario, Column A = Column B.
Case 2: Let \(\triangle ABC\) be a right triangle with \(\angle A = 30^{\circ}\) and \(\angle C = 60^{\circ}\).
\(\sin(A) = \sin(30^{\circ}) = \frac{1}{2}\).
\(\cos(C) = \cos(60^{\circ}) = \frac{1}{2}\).
In this scenario, Column A = Column B. (This occurs whenever A+C=90, since \(\sin(A) = \cos(90-A) = \cos(C)\)).
Case 3: Let's consider a non-right triangle. Let B=(0,3), D=(0,0).
Let A=(-4,0) and C=(1,0).
\(BD = 3\), \(AB = \sqrt{(-4-0)^2 + (0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5\).
\(DC = 1\), \(BC = \sqrt{(1-0)^2 + (0-3)^2} = \sqrt{1+9} = \sqrt{10}\).
Column A: \(\frac{BD}{AB} = \frac{3}{5} = 0.6\).
Column B: \(\frac{DC}{BC} = \frac{1}{\sqrt{10}} \approx \frac{1}{3.16} \approx 0.316\).
In this scenario, Column A \textgreater Column B.
Case 4: Let's switch the base lengths. B=(0,3), D=(0,0).
Let A=(-1,0) and C=(4,0).
\(BD = 3\), \(AB = \sqrt{(-1-0)^2 + (0-3)^2} = \sqrt{1+9} = \sqrt{10}\).
\(DC = 4\), \(BC = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5\).
Column A: \(\frac{BD}{AB} = \frac{3}{\sqrt{10}} \approx \frac{3}{3.16} \approx 0.949\).
Column B: \(\frac{DC}{BC} = \frac{4}{5} = 0.8\).
In this scenario, Column A \textgreater Column B. Let's try to make B \textgreater A. Case 5: Let B=(0,1), D=(0,0), A=(-10,0), C=(1,0). \(BD = 1\), \(AB = \sqrt{(-10)^2 + 1^2} = \sqrt{101}\). \(DC = 1\), \(BC = \sqrt{1^2 + 1^2} = \sqrt{2}\). Column A: \(\frac{BD}{AB} = \frac{1}{\sqrt{101}} \approx 0.1\). Column B: \(\frac{DC}{BC} = \frac{1}{\sqrt{2}} \approx 0.707\). In this scenario, Column B \textgreater Column A. Step 4: Final Answer:
Since we have found cases where A = B, A \textgreater B, and B \textgreater A, the relationship cannot be determined from the information given.