The question in the image reads as follows:

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 $\sqrt{3} = 1.73$).

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.

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.