To solve this problem, we'll use trigonometry, specifically the tangent function, which relates the angle of elevation to the height and distance of the object. Let's break down the steps and draw the diagram.

Step 1: Draw the Diagram

1. Draw a vertical line representing the flagpole with height $h$.
2. Mark two points on the ground along the same line of sight:
   • The first point $A$, where the student is initially standing, is a distance $x$ away from the base of the flagpole.
   • The second point $B$, where the student moves 2 meters further away, is a distance $x + 2$ from the base of the flagpole.
3. Draw the horizontal ground line connecting the base of the flagpole to both points $A$ and $B$.
4. Draw two lines from points $A$ and $B$ to the top of the flagpole, forming angles of elevation $70^\circ$ and $60^\circ$ respectively.

This will create two right triangles, $\triangle ABC$ and $\triangle ABD$, where:

• $h$ is the height of the flagpole,
• $x$ is the distance from the flagpole to point $A$,
• $x + 2$ is the distance from the flagpole to point $B$,
• $\theta_1 = 70^\circ$ (angle of elevation from point $A$),
• $\theta_2 = 60^\circ$ (angle of elevation from point $B$).

Step 2: Set Up Equations Using Tangent

For the first position (at point $A$): $\tan(70^\circ) = \frac{h}{x}$ So, $h = x \cdot \tan(70^\circ)$

For the second position (at point $B$): $\tan(60^\circ) = \frac{h}{x + 2}$ So, $h = (x + 2) \cdot \tan(60^\circ)$

Step 3: Set Up the Equation to Solve for $x$ and $h$

Since both expressions represent the height $h$, we can set them equal to each other: $x \cdot \tan(70^\circ) = (x + 2) \cdot \tan(60^\circ)$

Expanding and solving for $x$: $x \cdot \tan(70^\circ) = x \cdot \tan(60^\circ) + 2 \cdot \tan(60^\circ)$ $x \cdot (\tan(70^\circ) - \tan(60^\circ)) = 2 \cdot \tan(60^\circ)$ $x = \frac{2 \cdot \tan(60^\circ)}{\tan(70^\circ) - \tan(60^\circ)}$

Step 4: Calculate $x$ and then $h$

Substituting the values of the tangents: $\tan(70^\circ) \approx 2.747$ $\tan(60^\circ) \approx 1.732$

$x = \frac{2 \cdot 1.732}{2.747 - 1.732}$ $x \approx \frac{3.464}{1.015} \approx 3.413 \, \text{meters}$

Now, calculate $h$: $h = x \cdot \tan(70^\circ) \approx 3.413 \cdot 2.747 \approx 9.374 \, \text{meters}$

Final Answer:

The height of the flagpole is approximately 9.37 meters.

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Would you like further details on any part of the solution? Here are some related questions:

1. What other trigonometric identities could be used to solve similar problems?
2. How would the problem change if the student moved closer to the flagpole?
3. Can this method be applied to non-right triangles? How?
4. How does the accuracy of the angle measurements affect the result?
5. What if the ground was not level, how would that affect the calculations?

Tip: Always ensure that your calculator is set to the correct mode (degrees/radians) when solving trigonometric problems.