P and Q are two points observed from the top of a building `10sqrt(3)` m high. If the angles of depression of the points are complementary and PQ = 20 m, then the distance of P from the building is

Options

A. 30 m
B. 40 m
C. 25 m
D. 45 m

Answer

To solve the problem step by step, we will use the information provided in the question and the relationships between the angles and distances.

### Step 1: Understand the Setup
We have a building of height \( MO = 10\sqrt{3} \) m. Points P and Q are observed from the top of the building, and the angles of depression to these points are complementary. This means if the angle of depression to point P is \( \theta \), then the angle of depression to point Q is \( 90^\circ - \theta \).

### Step 2: Set Up the Triangles
Let:
- \( OP = x \) (the distance from the base of the building to point P)
- \( OQ = x + 20 \) (the distance from the base of the building to point Q, since PQ = 20 m)

Using the tangent function for the angles of depression:
1. For point P:
\[ \tan(\theta) = \frac{MO}{OP} = \frac{10\sqrt{3}}{x} \]
2. For point Q:
\[ \tan(90^\circ - \theta) = \cot(\theta) = \frac{MO}{OQ} = \frac{10\sqrt{3}}{x + 20} \]

### Step 3: Relate the Tangents
Since \( \tan(90^\circ - \theta) = \cot(\theta) \), we have:
\[ \cot(\theta) = \frac{1}{\tan(\theta)} \]
Substituting the values we found:
\[ \frac{10\sqrt{3}}{x + 20} = \frac{x}{10\sqrt{3}} \]

### Step 4: Cross Multiply
Cross multiplying gives us:
\[ (10\sqrt{3})^2 = x(x + 20) \]
Calculating \( (10\sqrt{3})^2 \):
\[ 300 = x^2 + 20x \]

### Step 5: Rearranging the Equation
Rearranging gives us the quadratic equation:
\[ x^2 + 20x - 300 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 20, c = -300 \):
\[ x = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-300)}}{2 \cdot 1} \]
\[ x = \frac{-20 \pm \sqrt{400 + 1200}}{2} \]
\[ x = \frac{-20 \pm \sqrt{1600}}{2} \]
\[ x = \frac{-20 \pm 40}{2} \]

This gives us two potential solutions:
1. \( x = \frac{20}{2} = 10 \)
2. \( x = \frac{-60}{2} = -30 \) (not valid since distance cannot be negative)

### Step 7: Find the Total Distance from the Building
Now that we have \( x = 10 \) m (the distance from the building to point Q), we can find the distance from the building to point P:
\[ OP = x = 10 \text{ m} \]
\[ OQ = x + 20 = 10 + 20 = 30 \text{ m} \]

### Final Answer
The distance of point P from the building is:
\[ \text{Distance of P from the building} = 10 \text{ m} \]