From the top of a platform 17 m high , the angle of elevation of the top of a tower was `30^(@)` . If the platform was positioned `50 sqrt3` m away from the tower , how tall was the tower ?

Options

A. `67` m
B. 50 m
C. `(25 sqrt3 + 17) `m
D. `25 sqrt3` m

Answer

To solve the problem step by step, we will use trigonometry to find the height of the tower.

### Step 1: Understand the given information
- Height of the platform (AB) = 17 m
- Distance from the platform to the tower (AD) = 50√3 m
- Angle of elevation (∠AED) = 30°

### Step 2: Set up the right triangle
From the top of the platform, we can visualize a right triangle where:
- AE is the height from the top of the platform to the top of the tower.
- AD is the horizontal distance from the platform to the base of the tower.
- DE is the vertical height of the tower above the platform.

### Step 3: Apply the tangent function
In triangle AED, we can use the tangent of the angle of elevation:
\[ \tan(30°) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{DE}{AD} \]
Where:
- DE is the height of the tower above the platform.
- AD = 50√3 m.

### Step 4: Substitute the known values
We know that:
\[ \tan(30°) = \frac{1}{\sqrt{3}} \]
So, substituting the values we have:
\[ \frac{1}{\sqrt{3}} = \frac{DE}{50\sqrt{3}} \]

### Step 5: Solve for DE
Cross-multiplying gives:
\[ DE = \frac{50\sqrt{3}}{\sqrt{3}} = 50 \text{ m} \]

### Step 6: Calculate the total height of the tower
The total height of the tower (EC) is the sum of the height of the platform (AB) and the height above the platform (DE):
\[ EC = AB + DE = 17 + 50 = 67 \text{ m} \]

### Final Answer
The height of the tower is **67 meters**.

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