From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are `45^(@)` and `60^(@)` respectively. Find the height of the tower. [Take `sqrt(3)=1.73`]

Answer

To solve the problem step by step, we will use the concepts of angles of depression and trigonometry.

### Step 1: Understand the Problem
We have a building that is 60 meters high, and we need to find the height of a tower based on the angles of depression from the top of the building to the top and bottom of the tower.

### Step 2: Draw a Diagram
- Let the height of the building be \( AB = 60 \) m.
- Let the height of the tower be \( CD \).
- Let the point where the bottom of the tower meets the ground be \( C \) and the top of the tower be \( D \).
- The angle of depression to the top of the tower \( D \) is \( 45^\circ \).
- The angle of depression to the bottom of the tower \( C \) is \( 60^\circ \).

### Step 3: Identify the Right Triangles
From the top of the building \( A \):
- Triangle \( ABE \) (where \( E \) is the foot of the perpendicular from \( A \) to the ground).
- Triangle \( ACD \) (where \( C \) is the bottom of the tower and \( D \) is the top of the tower).

### Step 4: Use Trigonometric Ratios
For triangle \( ABE \):
- The angle \( ABE = 45^\circ \).
- Using the tangent function:
\[ \tan(45^\circ) = \frac{AB}{BE} \implies 1 = \frac{60}{BE} \implies BE = 60 \text{ m} \]

For triangle \( ACD \):
- The angle \( ACD = 60^\circ \).
- Using the tangent function:
\[ \tan(60^\circ) = \frac{AB}{CE} \implies \sqrt{3} = \frac{60}{CE} \implies CE = \frac{60}{\sqrt{3}} = 20\sqrt{3} \text{ m} \]
(Using \( \sqrt{3} \approx 1.73 \)):
\[ CE \approx \frac{60}{1.73} \approx 34.64 \text{ m} \]

### Step 5: Find the Height of the Tower
The height of the tower \( CD \) can be found using:
\[ CD = AB - CE \]
Substituting the values:
\[ CD = 60 - 34.64 \approx 25.36 \text{ m} \]

### Final Answer
The height of the tower \( CD \) is approximately \( 25.36 \) meters.

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