The question involves a trigonometric problem that can be solved using tangent relationships and involves a tower's height and its shadow.

Problem:

• Given:

  1. The altitude of the sun changes from $45^\circ$ to $30^\circ$.
  2. The shadow increases by $10 \, \text{m}$.

• To Find: The height $h$ of the tower, correct to two decimal places.

Solution:

1. At $45^\circ$:

   • Using the tangent formula: $\tan(45^\circ) = \frac{\text{height of the tower (h)}}{\text{length of the shadow at } 45^\circ (x)}.$ Since $\tan(45^\circ) = 1$, we have: $h = x.$

2. At $30^\circ$:

   • Similarly, using the tangent formula: $\tan(30^\circ) = \frac{\text{height of the tower (h)}}{\text{length of the shadow at } 30^\circ (x + 10)}.$ Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have: $\frac{1}{\sqrt{3}} = \frac{h}{x + 10}.$

3. Substituting $h = x$ from the first equation into the second: $\frac{1}{\sqrt{3}} = \frac{x}{x + 10}.$

4. Cross-multiplying and simplifying: $x + 10 = \sqrt{3} \cdot x.$ $x(\sqrt{3} - 1) = 10.$ $x = \frac{10}{\sqrt{3} - 1}.$

5. Rationalizing the denominator: $x = \frac{10(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}.$ $x = \frac{10(\sqrt{3} + 1)}{2}.$ $x = 5(\sqrt{3} + 1).$

6. Calculating $x$: Using $\sqrt{3} \approx 1.732$: $x = 5(1.732 + 1) = 5 \cdot 2.732 = 13.66 \, \text{m}.$

7. Finding $h$: Since $h = x$, the height of the tower is: $h = 13.66 \, \text{m}.$

Final Answer:

The height of the tower is 13.66 meters.

Do you want further explanation or a breakdown of any step?

Related Questions:

1. What is the significance of rationalizing the denominator in such problems?
2. How do tangent functions relate to real-world problems like this?
3. Can you derive the formula for the shadow length as a function of the height and sun's angle?
4. What other angles apart from $30^\circ$ and $45^\circ$ could simplify the math?
5. How does changing the angle to $60^\circ$ instead affect the problem's solution?

Tip:

Always double-check tangent values and trigonometric simplifications in problems involving angles to avoid errors.