Given triangle $GHF$ inscribed in circle $c$, with points $H$ and $F$ on the circle and side lengths $GH$ = $a$, $HF$ = $b$, and $FG$ = $c$, what is the length of line segment $GH$ in terms of $b$, $c$, and the included angle $A$ opposite $GH$?

• A. $\sqrt{b^2-c^2+2bccosA}$
• B. $\sqrt{b^2+c^2-2bccosA}$
• C. $\sqrt{c^2+a^2-2accosB}$
• D. $\sqrt{b^2+c^2+2bccosA}$

Answer: (B)

Verified step by step guidance

1. Identify the triangle \( \triangle GHF \) with sides \( GH = a \), \( HF = b \), and \( FG = c \), and the angle \( A \) opposite side \( GH \).
2. Recall the Law of Cosines, which relates the lengths of sides of a triangle to the cosine of one of its angles: \[ a^2 = b^2 + c^2 - 2bc \cos A \]
3. Since \( a = GH \) is the side opposite angle \( A \), express \( a \) in terms of \( b \), \( c \), and \( A \) by taking the square root of both sides: \[ a = \sqrt{b^2 + c^2 - 2bc \cos A} \]
4. This formula gives the length of segment \( GH \) in terms of the other two sides \( b \) and \( c \), and the included angle \( A \).
5. Make sure to carefully substitute the known values of \( b \), \( c \), and \( A \) into the formula to find the length \( GH \) when needed.