Sure! Let's work through the problem step-by-step.
First, let's understand what we are given:
- The distance from the car to the foot of the building is [tex]\(46\)[/tex] meters.
- The height of the building is [tex]\(65\)[/tex] meters.
We are asked to find the angle of depression of the car from the top of the building. The angle of depression is the angle formed by the line of sight looking down from a point above (the top of the building in this case) to a point below (the car) with respect to the horizontal.
We can visualize this problem using a right triangle:
- The height of the building ([tex]\(65\)[/tex] meters) is the vertical leg of the triangle.
- The distance from the foot of the building to the car ([tex]\(46\)[/tex] meters) is the horizontal leg of the triangle.
- The angle of depression will be the same as the angle of elevation from the car to the top of the building as they are alternate interior angles.
To find the angle of depression, we will use the tangent function in trigonometry. The tangent of an angle in a right triangle is the ratio of the opposite leg (height of the building) to the adjacent leg (distance from the building).
The tangent function definition:
[tex]\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\][/tex]
Here, [tex]\(\theta\)[/tex] represents the angle of depression/elevation. Plugging in our values:
[tex]\[
\tan(\theta) = \frac{65}{46}
\][/tex]
Next, we need to find the angle [tex]\(\theta\)[/tex]. To do this, we take the arctangent (inverse tangent) of both sides:
[tex]\[
\theta = \arctan\left(\frac{65}{46}\right)
\][/tex]
Using this trigonometric function, the angle [tex]\(\theta\)[/tex] in radians can be calculated as approximately:
[tex]\[
\theta \approx 0.9549263748553701 \text{ radians}
\][/tex]
To convert this angle from radians to degrees, we multiply by [tex]\(\frac{180}{\pi}\)[/tex], because [tex]\(180\)[/tex] degrees is equivalent to [tex]\(\pi\)[/tex] radians:
[tex]\[
\theta_{\text{degrees}} = 0.9549263748553701 \times \frac{180}{\pi}
\][/tex]
This gives us:
[tex]\[
\theta_{\text{degrees}} \approx 54.713251024940284
\][/tex]
Therefore, the angle of depression of the car from the top of the building is approximately [tex]\(54.71\)[/tex] degrees.
