At night, a security camera pans over a parking lot. The camera is on a post at point a which is 53m from point c and 71 m from point b.the distance from B to c is 68m. Calculate the angle through which the camera pans
the law of cosines says
68^2 = 53^2 + 71^2 - 2*53*72 cosθ
so,
cosθ = (68^2-53^2-71^2)/(-2*53*72) = 0.42269
θ = 65°
Well, well, well, looks like we have a camera on a joyride! Let me help you calculate the angle of its wild pan.
Let's start by using the law of cosines to find the angle at point B. Here's the formula: c^2 = a^2 + b^2 - 2abcos(theta).
By plugging in the values we know, we get:
68^2 = 53^2 + 71^2 - 2(53)(71)cos(theta)
Now, let's solve for cos(theta):
cos(theta) = (53^2 + 71^2 - 68^2) / (2(53)(71))
cos(theta) = (2809 + 5041 - 4624) / (7546)
cos(theta) = 3226 / 7546
cos(theta) ≈ 0.427
To find the angle, we need to take the arccosine (inverse cosine) of 0.427. Grab your trusty calculator and find:
theta ≈ arccos(0.427)
theta ≈ 65.68 degrees
Voila! The angle through which the camera pans is approximately 65.68 degrees. That's quite a twist for the night watch, don't you think?
To find the angle through which the camera pans, we can use the Law of Cosines. Let's denote the angle at point B as θ. We can set up the equation as follows:
c² = a² + b² - 2ab * cos(θ)
Since we know the value of a, b, and c, we can plug in the values:
68² = 53² + 71² - 2 * 53 * 71 * cos(θ)
Simplifying:
4624 = 2809 + 5041 - 7526 * cos(θ)
Rearranging the equation:
7526 * cos(θ) = 2809 + 5041 - 4624
7526 * cos(θ) = 3226
Now, divide both sides of the equation by 7526:
cos(θ) = 3226 / 7526
θ = arccos(3226 / 7526)
Using a calculator, we find that the angle θ is approximately 57.2 degrees.
Therefore, the camera pans through an angle of approximately 57.2 degrees.
To calculate the angle through which the camera pans, we will use the Law of Cosines. This law states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides, minus twice their product multiplied by the cosine of the included angle.
In this case, we have a triangle with sides of lengths 53m, 71m, and 68m. Let's call the angle through which the camera pans θ.
Applying the Law of Cosines, we can set up the following equation:
68^2 = 53^2 + 71^2 - 2 * 53 * 71 * cos(θ)
Simplifying and rearranging the equation, we get:
4624 = 2809 + 5041 - 2 * 53 * 71 * cos(θ)
Rearranging further:
4624 - 2809 - 5041 = - 2 * 53 * 71 * cos(θ)
-3226 = - 2 * 53 * 71 * cos(θ)
Dividing by -2 * 53 * 71:
cos(θ) = -3226 / (-2 * 53 * 71)
cos(θ) = 3226 / (2 * 53 * 71)
cos(θ) = 0.955
To find the angle θ, we take the arccosine of 0.955:
θ = arccos(0.955)
Using a calculator or a math tool, we find that θ ≈ 17.6 degrees.
Therefore, the angle through which the camera pans is approximately 17.6 degrees.