Refer to the diagram shown below.
The height of the pole is CD (3 m), and its shadow is AB (1.22 m).
The height of the building is h m, and its shadow is AD (38.5 m).
The two triangles ΔABC and ΔADE are similar because of AAA.
Therefore
[tex] \frac{DE}{BC} = \frac{AD}{AB} \\\\ \frac{h}{3} = \frac{38.5}{1.22} [/tex]
Cross multiply.
1.22 h = 3*38.5 = 115.5
h = 115.5/1.22 = 140.91 m
Answer: 141 m (nearest meter)
