1. **Problem statement:** From a viewing tower 30 m above the ground, the angle of depression to an object on the ground is 36°. The angle of elevation to an aircraft vertically above the object is 48°. Find the distance of the aircraft from the ground. 2. **Diagram and variables:** Let the height of the tower be $AB = 30$ m. Let the object on the ground be point $C$. Let the aircraft be point $D$ vertically above $C$. We want to find $BD$, the distance from the ground to the aircraft. 3. **Using angle of depression:** Angle of depression from $A$ to $C$ is 36°, so angle $CAB = 36°$ (alternate interior angles). In right triangle $ABC$, $AB = 30$ m (height of tower), $BC$ is horizontal distance from tower base to object. Using tangent: $$\tan 36° = \frac{AB}{BC} = \frac{30}{BC} \implies BC = \frac{30}{\tan 36°}$$ Calculate $BC$: $$BC = \frac{30}{0.7265} \approx 41.3 \text{ m}$$ 4. **Using angle of elevation:** Angle of elevation from $C$ to $D$ is 48°. Since $D$ is vertically above $C$, $CD$ is vertical height of aircraft above ground. In right triangle $BCD$, $BC = 41.3$ m (horizontal), $\angle BCD = 48°$, and $BD$ is hypotenuse (distance from ground to aircraft). Using sine: $$\sin 48° = \frac{CD}{BD}$$ Using cosine: $$\cos 48° = \frac{BC}{BD}$$ We want $BD$, so: $$BD = \frac{BC}{\cos 48°} = \frac{41.3}{0.6691} \approx 61.7 \text{ m}$$ 5. **Final answer:** The distance of the aircraft from the ground is approximately **61.7 meters**.