Question

The angle of elevation of x from y is 30°,if
/xy/=40m how high is x above the level of y

Answer 1

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 \( 23.09 \, \text{m} \).