From the top of a tower of height 180 m the angles of depression of two objects on either sides of the tower are `30^(@)and45^(@)` . Then the distance between the objects are

Options

A. `180(3+sqrt(3))m`
B. `180(3-sqrt(3))`m
C. `180(sqrt(3)-1)`m
D. `180(sqrt(3)+1)`m

Answer

To solve the problem, let's break it down step by step.

### Step 1: Understand the Problem
We have a tower of height 180 m. From the top of the tower, the angles of depression to two objects on either side are 30° and 45°. We need to find the distance between these two objects.

### Step 2: Draw a Diagram
1. Draw a vertical line representing the tower (AB) with height 180 m.
2. Mark point A as the top of the tower and point B as the bottom.
3. From point A, draw two lines downwards at angles of 30° and 45° to represent the lines of sight to the objects (C and D).

### Step 3: Identify the Right Triangles
- For the object at angle 30° (let's call it point C), we can form a right triangle ABC.
- For the object at angle 45° (let's call it point D), we can form another right triangle ABD.

### Step 4: Use Trigonometric Ratios
1. **For angle 30°:**
- The tangent of 30° is given by:
\[ \tan(30°) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{180}{BC} \]
- We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{180}{BC} \]
- Rearranging gives:
\[ BC = 180 \sqrt{3} \]

2. **For angle 45°:**
- The tangent of 45° is given by:
\[ \tan(45°) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{180}{BD} \]
- We know that \(\tan(45°) = 1\), so:
\[ 1 = \frac{180}{BD} \]
- Rearranging gives:
\[ BD = 180 \]

### Step 5: Calculate the Total Distance Between the Objects
The total distance between the two objects (CD) is the sum of BC and BD:
\[ CD = BC + BD = 180\sqrt{3} + 180 \]
Factoring out 180 gives:
\[ CD = 180(\sqrt{3} + 1) \]

### Final Answer
The distance between the two objects is \(180(\sqrt{3} + 1)\) meters.

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