Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 122 ft from camera 2, which was 139 ft from camera 3. Cameras 1 and 3 were 151 ft apart. Which camera had to cover the greatest angle?(
Camera 1
Camera 2
Camera 3
There is not enough information to tell.
To determine which security camera had to cover the greatest angle, we can use the Law of Cosines, which allows us to find the angle opposite each side in a triangle when we know the lengths of all three sides.
The Law of Cosines states that for a triangle with sides of lengths \( a \), \( b \), and \( c \), and the angle opposite side \( c \) being \( \gamma \), the relationship between the sides and the angle is given by:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(\gamma) \]
We can solve for \( \cos(\gamma) \) to find the angle \( \gamma \) using the given side lengths.
Let's label the sides opposite each camera with the respective camera number:
- Side opposite Camera 1 (side length \( a \)): 139 ft
- Side opposite Camera 2 (side length \( b \)): 151 ft
- Side opposite Camera 3 (side length \( c \)): 122 ft
Using the Law of Cosines, we can find each angle:
For the angle \( A \) opposite Camera 1:
\[ b^2 + c^2 - a^2 = 2bc \cdot \cos(A) \]
\[ \cos(A) = \frac{151^2 + 122^2 - 139^2}{2 \cdot 151 \cdot 122} \]
For the angle \( B \) opposite Camera 2:
\[ a^2 + c^2 - b^2 = 2ac \cdot \cos(B) \]
\[ \cos(B) = \frac{139^2 + 122^2 - 151^2}{2 \cdot 139 \cdot 122} \]
For the angle \( C \) opposite Camera 3:
\[ a^2 + b^2 - c^2 = 2ab \cdot \cos(C) \]
\[ \cos(C) = \frac{139^2 + 151^2 - 122^2}{2 \cdot 139 \cdot 151} \]
Now let's calculate the cosine of each angle:
\[ \cos(A) = \frac{151^2 + 122^2 - 139^2}{2 \cdot 151 \cdot 122} \]
\[ \cos(A) = \frac{22801 + 14884 - 19321}{36862} \]
\[ \cos(A) = \frac{18364}{36862} \]
\[ \cos(B) = \frac{139^2 + 122^2 - 151^2}{2 \cdot 139 \cdot 122} \]
\[ \cos(B) = \frac{19321 + 14884 - 22801}{33938} \]
\[ \cos(B) = \frac{11404}{33938} \]
\[ \cos(C) = \frac{139^2 + 151^2 - 122^2}{2 \cdot 139 \cdot 151} \]
\[ \cos(C) = \frac{19321 + 22801 - 14884}{42089} \]
\[ \cos(C) = \frac{27238}{42089} \]
Since the cosine function is a decreasing function within the range of 0° to 180° (which covers the range of possible angles in a triangle), the smallest value of cosine corresponds to the largest angle. In our case, we're looking for the smallest value of the calculated cosines to identify the greatest angle.
By comparing the numerators (since the denominators are all positive), it's clear that:
\[ cos(B) = \frac{11404}{33938} \]
has the smallest numerator, therefore the smallest cosine value, which means that Camera 2 had to cover the greatest angle.