Question

In \(\triangle ABC, \angle B = 90^\circ\). If \( \frac{AB}{AC} = \frac{1}{2} \), then cos C is equal to

• A. \( \frac{3}{2} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( \frac{1}{\sqrt{3}} \)

Answer 1

The Correct Option is C

Given:
- In \(\triangle ABC\), \(\angle B = 90^\circ\) (right-angled at \(B\)).
- Ratio of sides: \(\frac{AB}{AC} = \frac{1}{2}\).
- Need to find \(\cos C\).

Step 1: Identify sides in the triangle
- Since \(\angle B = 90^\circ\), side \(AC\) is the hypotenuse.
- Sides \(AB\) and \(BC\) are the legs.

Step 2: Assign variables to sides
Let:
\[ AB = x \] \[ AC = 2x \quad (\text{from } \frac{AB}{AC} = \frac{1}{2}) \]

Step 3: Use Pythagoras theorem to find \(BC\)
\[ AC^2 = AB^2 + BC^2 \] Substitute values:
\[ (2x)^2 = x^2 + BC^2 \] \[ 4x^2 = x^2 + BC^2 \] \[ BC^2 = 4x^2 - x^2 = 3x^2 \] \[ BC = \sqrt{3}x \]

Step 4: Find \(\cos C\)
- \(\cos C = \frac{\text{adjacent side to } C}{\text{hypotenuse}} = \frac{BC}{AC}\)
\[ \cos C = \frac{\sqrt{3}x}{2x} = \frac{\sqrt{3}}{2} \]

Final Answer:
\[ \boxed{\frac{\sqrt{3}}{2}} \]