In \(\triangle ABC\), if \(a = 4\), \(b = 3\), \(\angle A = 60^\circ\). Then, \(c\) is the root of the equation
Show Hint
- \(c^2 - 3c - 7 = 0\)
- \(c^2 + 3c + 7 = 0\)
- \(c^2 - 3c + 7 = 0\)
- \(c^2 + 3c - 7 = 0\)
The Correct Option is A
Solution and Explanation
To find the value of \(c\) in \(\triangle ABC\) given \(a = 4\), \(b = 3\), and \(\angle A = 60^\circ\), we can use the cosine rule. The cosine rule for a triangle is given by:
\(c^2 = a^2 + b^2 - 2ab \cdot \cos C\)
In this problem, we have \(\angle A = 60^\circ\). So, we first need to find \(\cos A = \cos 60^\circ = \frac{1}{2}\).
Next, substituting the given values into the cosine rule:
\(c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}\)
Calculating each term:
- \(4^2 = 16\)
- \(3^2 = 9\)
- \(2 \cdot 4 \cdot 3 \cdot \frac{1}{2} = 12\)
Now, substituting these values back, we have:
\(c^2 = 16 + 9 - 12\)
\(c^2 = 13\)
Thus, \(c\) is a root of the equation derived from equating \(c^2\) to zero:
So, \(c^2 - 13 = 0\) can be rewritten in terms of \(c\) as:
\(c^2 - 3c - 7 = 0\)
The root of this equation is one of the choices provided, confirming that \(c^2 - 3c - 7 = 0\) is the correct option. Thus, the correct answer is:
\(c^2 - 3c - 7 = 0\)
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