The angular elevation of a tower OP at a point A due south of it is `60^@` and at a point B due west of A, the elevation is `30^@`. If AB=3m, the height of the tower is

Options

A. `2sqrt3`m
B. `2sqrt6`m
C. `(3sqrt3)/2` m
D. `(3sqrt6)/4` m

Explanation

To find the height of the tower OP, we can break down the problem step by step using trigonometric principles. Here’s the detailed solution:

### Step 1: Understand the Geometry
We have a tower OP, and two points A and B. Point A is due south of the tower, and point B is due west of point A. The angular elevation from point A to the top of the tower is \(60^\circ\), and from point B, it is \(30^\circ\). The distance between points A and B is given as \(AB = 3 \, \text{m}\).

### Step 2: Define Variables
Let:
- \(h\) = height of the tower OP
- \(OA\) = distance from point A to the base of the tower O
- \(OB\) = distance from point B to the base of the tower O

### Step 3: Use Trigonometric Ratios
From triangle AOP (where the angle of elevation is \(60^\circ\)):
\[ \tan(60^\circ) = \frac{h}{OA} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{h}{OA} \implies OA = \frac{h}{\sqrt{3}} \]

From triangle BOP (where the angle of elevation is \(30^\circ\)):
\[ \tan(30^\circ) = \frac{h}{OB} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we have:
\[ \frac{1}{\sqrt{3}} = \frac{h}{OB} \implies OB = h \sqrt{3} \]

### Step 4: Apply the Pythagorean Theorem
In triangle OAB, we know that:
\[ AB^2 = OA^2 + OB^2 \]
Substituting the values we found:
\[ 3^2 = \left(\frac{h}{\sqrt{3}}\right)^2 + (h \sqrt{3})^2 \]
This simplifies to:
\[ 9 = \frac{h^2}{3} + 3h^2 \]
Combining the terms:
\[ 9 = \frac{h^2}{3} + \frac{9h^2}{3} = \frac{10h^2}{3} \]

### Step 5: Solve for \(h^2\)
Multiplying both sides by 3:
\[ 27 = 10h^2 \implies h^2 = \frac{27}{10} \]

### Step 6: Find \(h\)
Taking the square root:
\[ h = \sqrt{\frac{27}{10}} = \frac{3\sqrt{3}}{\sqrt{10}} = \frac{3\sqrt{30}}{10} \]

### Final Answer
Thus, the height of the tower OP is:
\[ h = \frac{3\sqrt{30}}{10} \, \text{m} \]