The distance between the points `(5 cos 35^(@), 0)` and `(0, 5 cos 55^(@))` is

Options

A. 10 units
B. 5 units
C. 1 units
D. 2 units

Answer

To find the distance between the points \((5 \cos 35^\circ, 0)\) and \((0, 5 \cos 55^\circ)\), we will use the distance formula for two points in a Cartesian plane. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step-by-Step Solution:

1. **Identify the Points:**
- Let \(P_1 = (x_1, y_1) = (5 \cos 35^\circ, 0)\)
- Let \(P_2 = (x_2, y_2) = (0, 5 \cos 55^\circ)\)

2. **Substitute the Coordinates into the Distance Formula:**
\[ d = \sqrt{(0 - 5 \cos 35^\circ)^2 + (5 \cos 55^\circ - 0)^2} \]

3. **Simplify the Expression:**
\[ d = \sqrt{(-5 \cos 35^\circ)^2 + (5 \cos 55^\circ)^2} \]

4. **Calculate the Squares:**
\[ d = \sqrt{(25 \cos^2 35^\circ) + (25 \cos^2 55^\circ)} \]

5. **Factor Out the Common Term:**
\[ d = \sqrt{25 (\cos^2 35^\circ + \cos^2 55^\circ)} \]

6. **Use the Trigonometric Identity:**
- Recall that \( \cos^2 \theta + \sin^2 \theta = 1 \)
- We can express \(\cos^2 55^\circ\) as \(\sin^2(90^\circ - 55^\circ) = \sin^2 35^\circ\)

Therefore, we can use the identity:
\[ \cos^2 55^\circ = \sin^2 35^\circ \]

7. **Substitute the Identity:**
\[ d = \sqrt{25 (\cos^2 35^\circ + \sin^2 35^\circ)} \]

8. **Apply the Pythagorean Identity:**
\[ \cos^2 35^\circ + \sin^2 35^\circ = 1 \]
Thus,
\[ d = \sqrt{25 \cdot 1} = \sqrt{25} = 5 \]

### Final Answer:
The distance between the points \((5 \cos 35^\circ, 0)\) and \((0, 5 \cos 55^\circ)\) is \(5\) units.