1. **State the problem:** We have two right-angled triangles ABD and BCD sharing altitude BD perpendicular to AC. Given AD = 5 m, DC = 14 m, and angle BAD = 53°, we need to find the length BC. 2. **Identify what is needed:** BC is the sum of BD and DC, but since BD is vertical and DC is horizontal, we need to find BD and then use triangle BCD to find BC. 3. **Use trigonometry in triangle ABD:** - Angle BAD = 53° - AD = 5 m (adjacent side to angle BAD) - BD is opposite side to angle BAD Using tangent: $$\tan(53^\circ) = \frac{BD}{AD}$$ So, $$BD = AD \times \tan(53^\circ) = 5 \times \tan(53^\circ)$$ Calculate: $$\tan(53^\circ) \approx 1.3270$$ $$BD \approx 5 \times 1.3270 = 6.635$$ m 4. **Use Pythagoras theorem in triangle BCD:** - DC = 14 m (base) - BD = 6.635 m (height) Calculate BC: $$BC = \sqrt{BD^2 + DC^2} = \sqrt{6.635^2 + 14^2}$$ $$= \sqrt{44.02 + 196} = \sqrt{240.02} \approx 15.49$$ m **Final answer:** $$BC \approx 15.5$$ m (to 1 decimal place)