1. **Problem statement:** We have a vertical pole DC on horizontal ground ABC, where ABC is a straight line. The angle of elevation of point D from A is 27°. We know the distances AC = 22 m and BC = 7 m. We need to find the angle of elevation of D from B. 2. **Step 1: Calculate the height DC of the pole.** Using the right triangle ADC, where D is above C, and AC is the horizontal distance: The angle of elevation from A to D is 27°, so: $$\tan(27^\circ) = \frac{DC}{AC}$$ Rearranging to find DC: $$DC = AC \times \tan(27^\circ) = 22 \times \tan(27^\circ)$$ Calculate $\tan(27^\circ)$: approximately 0.5095. So: $$DC = 22 \times 0.5095 = 11.209$$ 3. **Step 2: Calculate the angle of elevation from B to D.** In triangle BDC, the horizontal distance BC = 7 m, and the vertical height DC = 11.209 m. The angle of elevation from B, call it $\theta$, satisfies: $$\tan(\theta) = \frac{DC}{BC} = \frac{11.209}{7}$$ Calculate the fraction: $$\frac{11.209}{7} = 1.6013$$ 4. **Step 3: Find $\theta$ by taking the arctangent:** $$\theta = \tan^{-1}(1.6013)$$ Using a calculator: $$\theta \approx 58.0^\circ$$ 5. **Final answer:** The angle of elevation of D from B is approximately **58°** to the nearest degree.