To solve the problem, we can use basic trigonometry. The angle of elevation of point \( x \) from point \( y \) is given as \( 30^\circ \), and the horizontal distance \( |xy| \) is \( 40 \) m.
We can visualize this scenario as a right triangle where:
- The horizontal distance \( |xy| \) is the adjacent side of the triangle.
- The vertical height \( h \) from point \( y \) to point \( x \) is the opposite side.
- The angle of elevation is \( 30^\circ \).
Using the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right triangle, we get:
\[ \tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} \]
Substituting the known values:
\[ \tan(30^\circ) = \frac{h}{40} \]
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{or} \quad \frac{\sqrt{3}}{3} \]
So we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{40} \]
Now, we can solve for \( h \):
\[ h = 40 \times \frac{1}{\sqrt{3}} = \frac{40}{\sqrt{3}} \]
To simplify this, we can multiply the numerator and denominator by \( \sqrt{3} \):
\[ h = \frac{40\sqrt{3}}{3} \]
Calculating \( \frac{40\sqrt{3}}{3} \):
\[ \approx \frac{40 \times 1.732}{3} \approx \frac{69.28}{3} \approx 23.09 , \text{m} \]
Therefore, the height \( x \) above the level of \( y \) is approximately \( 23.09 , \text{m} \).
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