1. **State the problem:** Kayla observes a plane flying at an altitude of 880 m. The angle of elevation changes from 67° 40' to 24° 30' after 25 seconds. We need to find (a) how far the plane has flown in that time and (b) the speed of the plane. 2. **Relevant formula:** For a right triangle, the tangent of the angle of elevation $\theta$ relates the altitude $h$ and horizontal distance $d$ by: $$\tan(\theta) = \frac{h}{d}$$ 3. **Convert angles to decimal degrees:** - $67^\circ 40' = 67 + \frac{40}{60} = 67.6667^\circ$ - $24^\circ 30' = 24 + \frac{30}{60} = 24.5^\circ$ 4. **Calculate horizontal distances at each angle:** - At $67.6667^\circ$: $$d_1 = \frac{h}{\tan(67.6667^\circ)} = \frac{880}{\tan(67.6667^\circ)}$$ - At $24.5^\circ$: $$d_2 = \frac{880}{\tan(24.5^\circ)}$$ 5. **Evaluate the tangents:** - $\tan(67.6667^\circ) \approx 2.43$ - $\tan(24.5^\circ) \approx 0.455$ 6. **Calculate distances:** $$d_1 = \frac{880}{2.43} \approx 362.14 \text{ m}$$ $$d_2 = \frac{880}{0.455} \approx 1934.07 \text{ m}$$ 7. **Calculate distance flown:** $$\text{Distance} = d_2 - d_1 = 1934.07 - 362.14 = 1571.93 \text{ m}$$ 8. **Calculate speed:** - Time = 25 seconds - Speed in m/s: $$v = \frac{1571.93}{25} = 62.88 \text{ m/s}$$ - Convert to km/h: $$v = 62.88 \times 3.6 = 226.37 \text{ km/h}$$ **Final answers:** - (a) The plane flew approximately 1572 m in 25 seconds. - (b) The speed of the plane is approximately 226 km/h.