Let's analyze the problem step by step.

1. Given:
   • The observer is standing on one bank of a river, looking at a tree on the opposite bank.
   • Initially, the angle of elevation to the top of the tree is $60^\circ$.
   • The observer moves 30 m away from the river, and the angle of elevation reduces to $30^\circ$.

To solve:

We can model this situation using trigonometry, particularly the concept of right triangles and the tangent function.

Question 1: What is the height of the tree?

• Let $h$ be the height of the tree, and let $x$ be the width of the river (the distance from the observer to the base of the tree).

  Using the tangent function: $\tan(60^\circ) = \frac{h}{x}$ $h = x \cdot \tan(60^\circ) = x \cdot \sqrt{3} \quad \text{(since } \tan(60^\circ) = \sqrt{3}\text{)}$

• After moving 30 m away from the river, the new distance from the observer to the tree is $x + 30$. The new angle of elevation is $30^\circ$, so: $\tan(30^\circ) = \frac{h}{x + 30}$ $\frac{1}{\sqrt{3}} = \frac{h}{x + 30}$ $h = \frac{x + 30}{\sqrt{3}}$

Now, equate the two expressions for $h$:

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

Now, substitute $x = 15$ into the first equation to find the height of the tree:

$h = 15 \cdot \sqrt{3} \approx 15 \cdot 1.732 = 25.98 \, \text{m}$ Thus, the height of the tree is approximately 25.98 m.

Question 2: What is the width of the river?

• From the previous calculations, the width of the river is 15 m.

Question 3: After moving how much distance away from the river does the angle of elevation become 45°?

• For the angle of elevation to be $45^\circ$, the distance from the tree must satisfy: $\tan(45^\circ) = \frac{h}{x + d}$ $1 = \frac{h}{x + d} \quad \text{(since } \tan(45^\circ) = 1\text{)}$ $h = x + d$ Substituting $h = 25.98$ m and $x = 15$ m: $25.98 = 15 + d$ $d = 25.98 - 15 = 10.98 \, \text{m}$ So, the observer must move 10.98 m away for the angle to become $45^\circ$.

Question 4: If the width of the river was 40 m, what would be the height of the tree?

• If $x = 40$ m, then using the first equation: $h = 40 \cdot \tan(60^\circ) = 40 \cdot \sqrt{3} \approx 40 \cdot 1.732 = 69.28 \, \text{m}$ The height of the tree would be 69.28 m.

Would you like more details on any of these solutions? Here are some related questions you might find helpful:

1. What if the observer had moved another 20 meters further from their final position?
2. How would the height of the tree change if the angles were $45^\circ$ and $15^\circ$?
3. Can you solve this using the sine rule?
4. How can the observer use only the 30-meter movement and angles to estimate tree height directly?
5. What if the tree was on the same side of the river but at a different height?

Tip: In trigonometry, angles of $30^\circ$, $45^\circ$, and $60^\circ$ often appear with simple trigonometric ratios, so it's helpful to memorize those!