Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Corrected text:

Law of cosines: [tex]$a^2 = b^2 + c^2 - 2bc \times \cos (A)$[/tex]


Sagot :

Certainly! Let's delve into the problem and solve it step-by-step using the Law of Cosines.

### Problem Statement:
You need to find the length of side [tex]\( a \)[/tex] in a triangle given the lengths of sides [tex]\( b \)[/tex] and [tex]\( c \)[/tex], and the angle [tex]\( A \)[/tex] between them.

The Law of Cosines states:
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]

where:
- [tex]\( a \)[/tex] is the side opposite angle [tex]\( A \)[/tex]
- [tex]\( b \)[/tex] and [tex]\( c \)[/tex] are the other two sides
- [tex]\( A \)[/tex] is the included angle between sides [tex]\( b \)[/tex] and [tex]\( c \)[/tex]

### Given Values:
- [tex]\( b = 0 \)[/tex]
- [tex]\( c = 0 \)[/tex]
- [tex]\( A = 0 \)[/tex] radians

### Steps to Solve for [tex]\( a \)[/tex]:

1. Substitute the given values into the Law of Cosines formula:
[tex]\[ a^2 = b^2 + c^2 - 2 \cdot b \cdot c \cdot \cos(A) \][/tex]
Plugging in the values, we get:
[tex]\[ a^2 = 0^2 + 0^2 - 2 \cdot 0 \cdot 0 \cdot \cos(0) \][/tex]

2. Simplify each term:
- [tex]\( 0^2 = 0 \)[/tex]
- Cosine of 0 radians, [tex]\(\cos(0)\)[/tex], is 1.

Therefore, the equation simplifies to:
[tex]\[ a^2 = 0 + 0 - 2 \cdot 0 \cdot 0 \cdot 1 \][/tex]
[tex]\[ a^2 = 0 - 0 \][/tex]
[tex]\[ a^2 = 0 \][/tex]

3. Take the square root of both sides to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \sqrt{0} \][/tex]
[tex]\[ a = 0 \][/tex]

### Conclusion:
The length of side [tex]\( a \)[/tex] is [tex]\( 0 \)[/tex].

In summary, after applying the Law of Cosines and substituting the given values, we find that the length of side [tex]\( a \)[/tex] is [tex]\( 0.0 \)[/tex].