18. In $\triangle A B C$ , the sides opposite to angles $A$ , $B$ , $C$ are $a$ , $b$ , $c$ respectively, with $b = a + 1$ , $c = a + 2$ . (1) If $2 \sin C = 3 \sin A$ , find the area of $\triangle A B C$ ; (2) Does there exist a positive integer $a$ such that $\triangle A B C$ is an obtuse triangle? If it exists, find the value of $a$ ; if not, explain the reason. Answer: (1) $\frac { 15 \sqrt { 7 } } { 4 }$ ; (2) Yes, and $a = 2$ .
[Solution]
[Analysis] (1) By the law of sines, we can obtain $2 c = 3 a$ . Combined with the known conditions, find the value of $a$ , and further obtain the values of $b$ and $c$ . Use the law of cosines and the fundamental trigonometric identity to find $\sin B$ , then use the area formula of a triangle to obtain the result; (2) Analyze that angle $C$ is obtuse. From $\cos C < 0$ combined with the triangle inequality, find the value of the integer $a$ . [Detailed Solution] (1) Since $2 \sin C = 3 \sin A$ , by the law of sines we have $2 c = 3 a$ . Thus $2 ( a + 2 ) = 3 a$ , so $a = 4$ . Therefore $b = 5$ , $c = 6$ . By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { 16 + 25 - 36 } { 2 \times 4 \times 5 } = \frac { 5 } { 40 } = \frac { 1 } { 8 }$ . Since $C$ is acute, $\sin C = \sqrt { 1 - \cos ^ { 2 } C } = \sqrt { 1 - \frac { 1 } { 64 } } = \frac { 3 \sqrt { 7 } } { 8 }$ . Therefore, $S _ { \triangle A B C } = \frac { 1 } { 2 } a b \sin C = \frac { 1 } { 2 } \times 4 \times 5 \times \frac { 3 \sqrt { 7 } } { 8 } = \frac { 15 \sqrt { 7 } } { 4 }$ ; (2) Clearly $c > b > a$ . If $\triangle A B C$ is an obtuse triangle, then $C$ is obtuse. By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { a ^ { 2 } + ( a + 1 ) ^ { 2 } - ( a + 2 ) ^ { 2 } } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } + a ^ 2 + 2a + 1 - a ^ 2 - 4a - 4 } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } - 2 a - 3 } { 2 a ( a + 1 ) } < 0$ , Solving: $- 1 < a < 3$ , thus $0 < a < 3$ . By the triangle inequality: $a + a + 1 > a + 2$ , we get $a > 1$ . Since $a \in \mathbb{Z}$ , we have $a = 2$ .
18. In $\triangle A B C$ , the sides opposite to angles $A$ , $B$ , $C$ are $a$ , $b$ , $c$ respectively, with $b = a + 1$ , $c = a + 2$ .\\
(1) If $2 \sin C = 3 \sin A$ , find the area of $\triangle A B C$ ;\\
(2) Does there exist a positive integer $a$ such that $\triangle A B C$ is an obtuse triangle? If it exists, find the value of $a$ ; if not, explain the reason.
\textbf{Answer:} (1) $\frac { 15 \sqrt { 7 } } { 4 }$ ; (2) Yes, and $a = 2$ .
\section*{[Solution]}
[Analysis] (1) By the law of sines, we can obtain $2 c = 3 a$ . Combined with the known conditions, find the value of $a$ , and further obtain the values of $b$ and $c$ . Use the law of cosines and the fundamental trigonometric identity to find $\sin B$ , then use the area formula of a triangle to obtain the result;\\
(2) Analyze that angle $C$ is obtuse. From $\cos C < 0$ combined with the triangle inequality, find the value of the integer $a$ .\\
[Detailed Solution] (1) Since $2 \sin C = 3 \sin A$ , by the law of sines we have $2 c = 3 a$ . Thus $2 ( a + 2 ) = 3 a$ , so $a = 4$ . Therefore $b = 5$ , $c = 6$ . By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { 16 + 25 - 36 } { 2 \times 4 \times 5 } = \frac { 5 } { 40 } = \frac { 1 } { 8 }$ . Since $C$ is acute, $\sin C = \sqrt { 1 - \cos ^ { 2 } C } = \sqrt { 1 - \frac { 1 } { 64 } } = \frac { 3 \sqrt { 7 } } { 8 }$ .
Therefore, $S _ { \triangle A B C } = \frac { 1 } { 2 } a b \sin C = \frac { 1 } { 2 } \times 4 \times 5 \times \frac { 3 \sqrt { 7 } } { 8 } = \frac { 15 \sqrt { 7 } } { 4 }$ ;\\
(2) Clearly $c > b > a$ . If $\triangle A B C$ is an obtuse triangle, then $C$ is obtuse.\\
By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { a ^ { 2 } + ( a + 1 ) ^ { 2 } - ( a + 2 ) ^ { 2 } } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } + a ^ 2 + 2a + 1 - a ^ 2 - 4a - 4 } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } - 2 a - 3 } { 2 a ( a + 1 ) } < 0$ ,\\
Solving: $- 1 < a < 3$ , thus $0 < a < 3$ .\\
By the triangle inequality: $a + a + 1 > a + 2$ , we get $a > 1$ . Since $a \in \mathbb{Z}$ , we have $a = 2$ .\\