The problem in the image involves using trigonometry to calculate the height of the tree and the width of the river, based on the observer's angles of elevation from different distances. Let's break it down step-by-step and address each question.

Given:

1. When the observer stands on the riverbank, the angle of elevation to the top of the tree is $60^\circ$.
2. After moving 30 m away from the river, the angle of elevation reduces to $30^\circ$.

Using trigonometry, we can solve the following:

1. What is the height of the tree?

Let's denote:

• $h$ as the height of the tree.
• $d$ as the width of the river (the horizontal distance from the observer at the initial position to the base of the tree).

From the observer's initial position: $\tan(60^\circ) = \frac{h}{d}$ We know that $\tan(60^\circ) = \sqrt{3}$, so: $\sqrt{3} = \frac{h}{d}$ $h = d\sqrt{3}$

2. What is the width of the river?

Now, after the observer moves 30 m away from the river, the new horizontal distance becomes $d + 30$ m. The new angle of elevation is $30^\circ$: $\tan(30^\circ) = \frac{h}{d + 30}$ We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, so: $\frac{1}{\sqrt{3}} = \frac{h}{d + 30}$ $h = \frac{d + 30}{\sqrt{3}}$

Substituting $h = d\sqrt{3}$ into the second equation:

$d\sqrt{3} = \frac{d + 30}{\sqrt{3}}$ Multiply both sides by $\sqrt{3}$ to eliminate the denominator: $3d = d + 30$ $2d = 30$ $d = 15 \, \text{m}$ So, the width of the river is $15 \, \text{m}$.

1. What is the height of the tree?

Substitute $d = 15$ m into $h = d\sqrt{3}$: $h = 15\sqrt{3} \approx 15 \times 1.732 = 25.98 \, \text{m}$ Thus, the height of the tree is approximately $25.98 \, \text{m}$.

3. After moving how much distance away from the river does the angle of elevation become $45^\circ$?

For the angle to be $45^\circ$, the horizontal distance from the tree base should equal the tree height, because: $\tan(45^\circ) = 1 \quad \Rightarrow \quad \frac{h}{d + x} = 1$ Thus, the total distance from the tree will be $h$, which we calculated as $25.98$ m. Since the width of the river is $15$ m, the extra distance moved $x$ can be found as: $d + x = h \quad \Rightarrow \quad 15 + x = 25.98$ $x = 25.98 - 15 = 10.98 \, \text{m}$ So, after moving approximately $10.98 \, \text{m}$, the angle of elevation becomes $45^\circ$.

4. If the width of the river were 40 m, what would be the height of the tree?

For $d = 40 \, \text{m}$: $\tan(60^\circ) = \frac{h}{40} \quad \Rightarrow \quad \sqrt{3} = \frac{h}{40}$ $h = 40\sqrt{3} \approx 40 \times 1.732 = 69.28 \, \text{m}$ Thus, the height of the tree would be approximately $69.28 \, \text{m}$ if the river were 40 m wide.

Would you like more details on any of the steps, or do you have any questions?

Here are 5 related questions to explore further:

1. How would the solution change if the initial angle of elevation was $45^\circ$?
2. What would happen to the height if the observer moved twice the distance away from the riverbank?
3. How can you verify the correctness of the trigonometric values used?
4. How can you use these calculations to estimate the height of a real object in nature?
5. What other real-world applications can you find for this type of trigonometry problem?

Tip: Always ensure the angle of elevation is measured accurately from the horizontal line when solving these problems in real-life settings.