The intersections D and C are DE and CF perpendicular to AB, respectively, and the vertical legs are E and F. 。
∴DE=CF=AC×cos 12 ≈29.3。
Therefore, the distanc
The intersections D and C are DE and CF perpendicular to AB, respectively, and the vertical legs are E and F. 。
∴DE=CF=AC×cos 12 ≈29.3。
Therefore, the distance from point D to tree AB is 29.3 meters.
∫AF = AC×sin 12，BE=BE×tan 10。
The height of tree AB is BE+EF+FA≈ 12.9 meters.