\(5\sqrt{2+\sqrt{2}}\) km
To find the height of the plane above the ground given the angle of elevation and the distance from the point of observation, we'll use trigonometric relationships. In this scenario, the problem involves calculating the height of an object using the tangent of the angle of elevation.
Given:
- Distance from the point of observation (horizontal distance) \( = 10 \text{ km} \)
- Angle of elevation \( = 67.5^\circ \)
The tangent of an angle in a right triangle is the ratio of the opposite side (height of the object) to the adjacent side (horizontal distance). The formula is:
\(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\)
In this problem, \(\theta = 67.5^\circ\), the opposite side is the height of the plane \( h \), and the adjacent side is the horizontal distance 10 km.
Thus, we have:
\(\tan(67.5^\circ) = \frac{h}{10}\)
To find the value of \(h\), rearrange the formula:
\(h = 10 \times \tan(67.5^\circ)\)
Let us calculate \(\tan(67.5^\circ)\). The angle \( 67.5^\circ \) is related to a known trigonometric identity:
\(\tan(67.5^\circ) = \tan\left(45^\circ + 22.5^\circ\right)\)
Using the tangent addition formula:
\(\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}\)
For \(a = 45^\circ\) and \(b = 22.5^\circ\), knowing \(\tan(45^\circ) = 1\) and using identities, we can solve for exact trigonometric values at special angles:
We use the derived result for this common angle:
\(\tan(67.5^\circ) = 2 + \sqrt{2}\)
Substitute back into the height equation:
\(h = 10 \times (2 + \sqrt{2})\)
Therefore, after simplifying, the height \(h\) is:
\(h = 5\sqrt{2+\sqrt{2}} \, \text{km}\)
Thus, the correct answer is \(h = 5\sqrt{2+\sqrt{2}} \, \text{km}\). This matches the given option.
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