A man stands on the ground at a point A, Which is on the same horizontal plane as B, the foot of a vertical Pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground He observes the angle of clevation at C, the top of the pole as $\displaystyle x^{\circ}$, where tan $\displaystyle x^{\circ}=\frac{2}{5}.$ The distance AB ( in meters ) is

The vertical distance from the man's eye level to the top of the pole is 10 meters minus 2 meters, which equals 8 meters. The angle of elevation satisfies tan(x degrees) equals the opposite height divided by the adjacent distance AB, so 2 divided by 5 equals 8 divided by the distance AB. Cross-multiplying gives 2 times the distance AB equals 40, meaning the distance AB is 20 meters.

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