A ladder is placed along a wall such that its upper end is touching the top of the wall .The foot of the ladder is 10 ft away from the wall and the ladder is making an angle of `60^(@)` with the ground . When a man starts climbing on it , it slips and now ladder makes an angle of `30^(@)` with ground .How much did the ladder slip from the of the wall ?

Options

A. 12 ft
B. 20 ft
C. 7.32 ft
D. 18 ft

Answer

To solve the problem step by step, we will analyze the situation involving the ladder, the wall, and the angles formed.

### Step 1: Understand the initial setup
We have a ladder (BC) leaning against a wall (AB). The foot of the ladder (C) is 10 ft away from the wall (A). The ladder makes an angle of 60° with the ground.

### Step 2: Identify the lengths and angles
- Let AC = 10 ft (the distance from the wall to the foot of the ladder).
- Let BC = L (the length of the ladder).
- The angle ∠BCA = 60°.

### Step 3: Use trigonometry to find the length of the ladder
Using the cosine function:
\[ \cos(60°) = \frac{AC}{BC} \]
Substituting the known values:
\[ \cos(60°) = \frac{10}{L} \]
Since \(\cos(60°) = \frac{1}{2}\):
\[ \frac{1}{2} = \frac{10}{L} \]
Cross-multiplying gives:
\[ L = 20 \text{ ft} \]

### Step 4: Set up the scenario after the ladder slips
After the man climbs the ladder, it slips and now makes an angle of 30° with the ground. Let CD be the distance the ladder slips from the wall, which we will denote as x ft.

### Step 5: Identify the new position of the foot of the ladder
Now, the distance from the wall to the foot of the ladder (AD) is:
\[ AD = AC + CD = 10 + x \]

### Step 6: Use trigonometry for the new angle
Using the cosine function again for the new angle:
\[ \cos(30°) = \frac{AD}{BC} \]
Substituting the known values:
\[ \cos(30°) = \frac{10 + x}{20} \]
Since \(\cos(30°) = \frac{\sqrt{3}}{2}\):
\[ \frac{\sqrt{3}}{2} = \frac{10 + x}{20} \]
Cross-multiplying gives:
\[ 20\sqrt{3} = 2(10 + x) \]
Simplifying:
\[ 20\sqrt{3} = 20 + 2x \]

### Step 7: Solve for x
Rearranging the equation:
\[ 2x = 20\sqrt{3} - 20 \]
Dividing by 2:
\[ x = 10\sqrt{3} - 10 \]

### Step 8: Calculate the numerical value of x
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ x \approx 10(1.732) - 10 = 17.32 - 10 = 7.32 \text{ ft} \]

### Final Answer
The ladder slipped approximately **7.32 ft** from the wall.

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