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The question in the image reads as follows:

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**35.** The angle of elevation of the top \( Q \) of a vertical tower \( PQ \) from a point \( X \) on the ground is \( 60^\circ \). At a point \( Y \), 40 m vertically above \( X \), the angle of elevation is \( 45^\circ \). Find the height of the tower \( PQ \).  
(Take \( \sqrt{3} = 1.73 \)).

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### Let's solve this problem step by step.

Let:
- The height of the tower \( PQ = h \).
- The horizontal distance from \( X \) to the base of the tower \( P \) be \( d \).

1. From point \( X \), using the angle of elevation of \( 60^\circ \):

\[
\tan(60^\circ) = \frac{h}{d}
\]

Since \( \tan(60^\circ) = \sqrt{3} = 1.73 \), we have:

\[
1.73 = \frac{h}{d} \implies h = 1.73d
\]

2. From point \( Y \), 40 m above \( X \), the height above ground is \( h - 40 \). Using the angle of elevation of \( 45^\circ \):

\[
\tan(45^\circ) = \frac{h - 40}{d}
\]

Since \( \tan(45^\circ) = 1 \), we have:

\[
1 = \frac{h - 40}{d} \implies h - 40 = d \implies d = h - 40
\]

3. Now we have two equations:

\[
h = 1.73d
\]
\[
d = h - 40
\]

Substitute \( d = h - 40 \) into \( h = 1.73d \):

\[
h = 1.73(h - 40)
\]
\[
h = 1.73h - 69.2
\]
\[
69.2 = 0.73h
\]
\[
h = \frac{69.2}{0.73} \approx 94.79 \, \text{m}
\]

So, the height of the tower \( PQ \) is approximately **94.79 meters**.

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Would you like further clarifications or details on this? Here are 5 related questions to expand your understanding:

1. How would the calculation change if the angle of elevation from \( X \) was different, say \( 45^\circ \)?
2. What if the point \( Y \) was located 50 m above \( X \) instead of 40 m?
3. How can you verify the solution using trigonometric identities?
4. How would the problem change if the distance from \( Y \) to the base was known?
5. What is the significance of knowing the tangent values for specific angles in real-life problems?

**Tip**: For angle of elevation problems, always focus on setting up correct trigonometric ratios using the given angles and distances.

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of the tower PQ. [Take √3 = 1.73]

Learn how to calculate the height of a tower using trigonometric ratios from angles of elevation at 60° and 45°. This example covers key trigonometry concepts and is suitable for students in Grades 10-12.

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