From the top of a 10 m high building, the angle of elevation of the top of a tower is `60^@` and the angle of depression of the foot of the tower is `Theta`, such that `tan Theta = 2/3`. What is the height of the tower to nearest metres?
10 मी ऊँची इमारत से किसी मीनार के शीर्ष का उन्नयन कोण `60^@` है तथा मीनार के तल का अवनमन कोण `Theta` इस प्रकार है कि`tan Theta = 2/3` है | निकटतम मीटर तक मीनार की ऊंचाई ज्ञात करें |

Options

A. 34m
B. 35 m
C. 36m
D. 33m

Answer

To solve the problem, we will break it down into steps:

### Step 1: Understand the scenario
We have a building that is 10 meters high. From the top of this building, we observe a tower. The angle of elevation to the top of the tower is \(60^\circ\) and the angle of depression to the foot of the tower is \(\Theta\), where \(\tan \Theta = \frac{2}{3}\).

### Step 2: Determine the height of the tower
Let’s denote:
- The height of the tower as \(h\).
- The distance from the base of the building to the foot of the tower as \(d\).

From the top of the building, we can create two right triangles:
1. The triangle formed by the height of the building, the height of the tower, and the distance \(d\) (for the angle of elevation).
2. The triangle formed by the height of the building, the height of the tower, and the distance \(d\) (for the angle of depression).

### Step 3: Use the angle of elevation
Using the angle of elevation \(60^\circ\):
\[ \tan(60^\circ) = \frac{h - 10}{d} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{h - 10}{d} \quad \Rightarrow \quad h - 10 = d \sqrt{3} \quad \Rightarrow \quad h = d \sqrt{3} + 10 \]

### Step 4: Use the angle of depression
Using the angle of depression \(\Theta\):
\[ \tan(\Theta) = \frac{10}{d} \]
Given that \(\tan \Theta = \frac{2}{3}\), we have:
\[ \frac{2}{3} = \frac{10}{d} \quad \Rightarrow \quad d = \frac{10 \cdot 3}{2} = 15 \]

### Step 5: Substitute \(d\) back to find \(h\)
Now substituting \(d = 15\) back into the equation for \(h\):
\[ h = 15\sqrt{3} + 10 \]

### Step 6: Calculate \(h\)
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ h = 15 \cdot 1.732 + 10 = 25.98 + 10 = 35.98 \]

### Step 7: Round to the nearest meter
Rounding \(35.98\) to the nearest meter gives us:
\[ h \approx 36 \text{ meters} \]

### Final Answer
The height of the tower is approximately **36 meters**.

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