Let's draw and label the situation: - Let **T** be Thea's position on the ground. - Let **E** be Ella's position on the ground, **6 feet closer** to the kite than Thea. - Let **K** be the position of the kite in the air, directly above the ground (not necessarily above either girl). - The ground is level. Let: - \( h \) = **vertical height** of the kite above the ground (this is what we want) - \( d \) = **horizontal distance** from Thea to the point on the ground directly below the kite Given: - The angle from Thea's feet to the kite is \( 30^\circ \). - The angle from Ella's feet to the kite is \( 45^\circ \). - Ella is 6 feet closer to the kite horizontally than Thea, i.e., **the horizontal distance from Ella to the kite is \( d - 6 \)**. **Let's set up the equations using right triangle trigonometry.** --- ## For Thea: \[ \tan(30^\circ) = \frac{h}{d} \] \[ h = d \cdot \tan(30^\circ) \] --- ## For Ella: \[ \tan(45^\circ) = \frac{h}{d - 6} \] \[ h = (d - 6) \cdot \tan(45^\circ) \] But \(\tan(45^\circ) = 1\), so: \[ h = d - 6 \] --- ## Set the two expressions for \( h \) equal: \[ d \cdot \tan(30^\circ) = d - 6 \] \[ d \cdot \tan(30^\circ) - d = -6 \] \[ d(\tan(30^\circ) - 1) = -6 \] \[ d = \frac{-6}{\tan(30^\circ) - 1} \] --- ## Calculate \( \tan(30^\circ) \): \[ \tan(30^\circ) = \frac{\sqrt{3}}{3} \approx 0.5774 \] So, \[ d = \frac{-6}{0.5774 - 1} = \frac{-6}{-0.4226} \] \[ d \approx \frac{6}{0.4226} \approx 14.2 \text{ feet} \] --- ## Now, find \( h \): \[ h = d \cdot \tan(30^\circ) \approx 14.2 \cdot 0.5774 \approx 8.2 \text{ feet} \] --- ### **Final Answer:** \[ \boxed{8.2\ \text{feet}} \] The vertical height of the kite from the ground is **about 8.2 feet**.

Let's carefully analyze the problem step-by-step: **Given:** - Thea and Ella are on level ground, both to the left of the kite. - Ella is 6 feet closer to the kite than Thea. - The angle of elevation from Thea's feet to the kite is 30°. - The angle of elevation from Ella's feet to the kite is 45°. - Goal: Find the vertical height \( h \) of the kite above ground. --- ### Step 1: Define variables - \( d_T \): horizontal distance from Thea to the point directly beneath the kite. - \( d_E \): horizontal distance from Ella to the point directly beneath the kite. - Since Ella is 6 feet closer to the kite than Thea: \[ d_E = d_T - 6 \] - \( h \): vertical height of the kite (unknown). --- ### Step 2: Write equations using tangent of angles **For Thea:** \[ \tan(30^\circ) = \frac{h}{d_T} \Rightarrow h = d_T \cdot \tan(30^\circ) \] **For Ella:** \[ \tan(45^\circ) = \frac{h}{d_E} \Rightarrow h = d_E \cdot \tan(45^\circ) \] Since \(\tan(45^\circ) = 1\), this simplifies to: \[ h = d_E \] But recall: \[ d_E = d_T - 6 \] So: \[ h = d_T - 6 \] --- ### Step 3: Set the two expressions for \( h \) equal \[ d_T \cdot \tan(30^\circ) = d_T - 6 \] Substitute \(\tan(30^\circ) = \frac{\sqrt{3}}{3} \approx 0.5774\): \[ d_T \times 0.5774 = d_T - 6 \] Rearranged: \[ 0.5774 d_T - d_T = -6 \] \[ (-0.4226) d_T = -6 \] \[ d_T = \frac{-6}{-0.4226} \approx 14.2 \text{ feet} \] --- ### Step 4: Find the height \( h \) \[ h = d_T \times \tan(30^\circ) \approx 14.2 \times 0.5774 \approx 8.2 \text{ feet} \] --- ### **Final answer:** \[ \boxed{\text{The height of the kite is approximately } \mathbf{8.2 \text{ feet}}} \] This completes the solution.