Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m, and CA = 6 m. A vertical lamp-post at the midpoint D of AC subtends an angle 30°. The height (in m) of the lamp-post is?

Let h be the height of the lamp-post. In triangle ABC, let a = BC = 5, b = AC = 6, c = AB = 7. The semi-perimeter s is given by s = (a+b+c)/2 = (5+6+7)/2 = 9. Using Heron's formula, the area of triangle ABC is √(s(s-a)(s-b)(s-c)) = √(9(9-5)(9-6)(9-7)) = √(943*2) = √216 = 6√6.

In triangle ABC, the area is also given by (1/2) * base * height = (1/2) * AC * h_ABC = (1/2) * 6 * h_ABC = 3h_ABC, where h_ABC is the height from B to AC.

Therefore, 3h_ABC = 6√6, which gives h_ABC = 2√6.

Now consider the triangle BDC. D is the midpoint of AC, so AD = DC = 3. Let the angle BDC be θ = 30°. In triangle BDC, we can use the tangent formula:

tan(30°) = h / (BD)

1/√3 = h / BD

BD = h√3

By Apollonius' theorem in triangle ABC:

AB² + BC² = 2(BD² + AD²)

7² + 5² = 2(BD² + 3²)

49 + 25 = 2(BD² + 9)

74 = 2BD² + 18

2BD² = 56

BD² = 28

BD = √28 = 2√7

Substituting BD = h√3:

h√3 = 2√7

h = 2√7 / √3 = 2√(21)/3

However, this solution uses Apollonius' theorem and trigonometric functions which are not directly used in the given solution. Let's try another approach.

In triangle BCD, we have BD = h cot(30°) = h√3. By the cosine rule in triangle ABC:

BC² = AB² + AC² - 2(AB)(AC)cos(BAC)

25 = 49 + 36 - 2(7)(6)cos(BAC)

84cos(BAC) = 60

cos(BAC) = 60/84 = 5/7

Area of triangle ABC = (1/2)(7)(6)sin(BAC) = 21√(1 - (5/7)²) = 21√(24/49) = 6√6

Area of triangle ABC = (1/2)(AC)(h) = 3h

3h = 6√6

h = 2√6

There seems to be an error in the provided solution. The correct calculation using the Apollonius theorem is shown above. The final solution using the trigonometric approach with the proper area calculation leads to h = 2√6. None of the options match this result. There might be an error in the question or the options provided.