15. As shown in the figure, a car is traveling due west on a horizontal road. At point A, the mountain peak D on the north side of the road is measured to be in the direction of $30 ^ { 0 }$ west of north. After traveling 600 m to reach point B, the mountain peak is measured to be in the direction of $75 ^ { 0 }$ west of north, with an angle of elevation of $30 ^ { 0 }$. Then the height of the mountain $\mathrm { CD } =$ $\_\_\_\_$ m. [Figure]

15. In $\triangle ABC$, given $A B = 2 , A C = 3 , A = 60 ^ { \circ }$ .
(1) Find the length of BC;
(2) Find the value of $\sin 2 C$.

16. (13 points) In $\triangle ABC$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. Given that the area of $\triangle ABC$ is $3 \sqrt { 15 }$, $b - c = 2$, $\cos A = - \frac { 1 } { 4 }$. (I) Find the values of $a$ and $\sin C$; (II) Find the value of $\cos \left( 2 A + \frac { \pi } { 6 } \right)$.

16. (This question is worth 14 points) In $\triangle ABC$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given that $A = \frac { \pi } { 4 }$ and $b ^ { 2 } - a ^ { 2 } = \frac { 1 } { 2 } c ^ { 2 }$ . (I) Find the value of $\tan C$; (II) If the area of $\triangle ABC$ is 7, find the value of $b$.

In $\triangle \mathrm { ABC }$, $D$ is a point on $BC$, $AD$ bisects $\angle \mathrm { BAC }$, and the area of $\triangle \mathrm { ABD }$ is 2 times the area of $\triangle \mathrm { ADC }$.
(I) Find $\frac { \sin \angle B } { \sin \angle C }$ ;
(II) If $A D = 1 , D C = \frac { \sqrt { 2 } } { 2 }$, find the lengths of $B D$ and $A C$.

17. (12 points) In $\triangle \mathrm { ABC }$, the sides opposite to angles $\mathrm { A } , \mathrm { B }$, C are $a , b , c$ respectively. The vector $\vec { m } = ( a , \sqrt { 3 } b )$ is parallel to $\vec { n } = ( \cos \mathrm { A } , \sin \mathrm { B } )$. (I) Find A; (II) If $a = \sqrt { 7 } , b = 2$, find the area of $\triangle \mathrm { ABC }$.

18. (This question is worth 12 points). A company randomly surveyed 40 users from regions $A$ and $B$ respectively to understand user satisfaction with its products. Based on user satisfaction scores, a frequency distribution histogram of user satisfaction scores in region $A$ and a frequency distribution table of user satisfaction scores in region $B$ were obtained.
Frequency Distribution Histogram of User Satisfaction Scores in Region A [Figure]

Satisfaction Score Interval	\multicolumn{5}{|c|}{Frequency Distribution Table of User Satisfaction Scores in Region B}
Frequency	$[ 50,60 )$	$[ 60,70 )$	$[ 70,80 )$	$[ 80,90 )$	$[ 90,100 ]$
	2	8	14	10	6

(I) Draw the frequency distribution histogram of user satisfaction scores in region $B$ on the answer sheet, and compare the mean and dispersion of satisfaction scores in the two regions through this histogram (no need to calculate specific values, just provide conclusions); [Figure] (II) Based on user satisfaction scores, user satisfaction levels are divided into three categories:

Satisfaction Score	Below 70	70 to 89	At least 90
Satisfaction Level	Dissatisfied	Satisfied	Very Satisfied

Estimate which region has a higher probability that a user's satisfaction level is dissatisfied, and explain the reasoning.

In $\triangle ABC$, the interior angles $A$, $B$, $C$ have opposite sides $a$, $b$, $c$ respectively. If $2b\cos B = a\cos C + c\cos A$, then $B = $ \_\_\_\_

In triangle $ABC$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. Given that the area of $\triangle ABC$ is $\frac { a ^ { 2 } } { 3 \sin A }$.
(1) Find $\sin B \sin C$;
(2) If $b + c = 2$, find the range of values of $a$.

In $\triangle A B C$, $\cos \frac { C } { 2 } = \frac { \sqrt { 5 } } { 5 } , B C = 1 , A C = 5$, then $A B =$
A. $4 \sqrt { 2 }$
B. $\sqrt { 30 }$
C. $\sqrt { 29 }$
D. $2 \sqrt { 5 }$

In $\triangle A B C$, $\cos \frac { C } { 2 } = \frac { \sqrt { 5 } } { 5 } , B C = 1 , A C = 5$, then $A B =$
A. $4 \sqrt { 2 }$
B. $\sqrt { 30 }$
C. $\sqrt { 29 }$
D. $2 \sqrt { 5 }$

In $\triangle ABC$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If the area of $\triangle ABC$ equals $\frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 4 }$, then $C =$
A. $\frac { \pi } { 2 }$
B. $\frac { \pi } { 3 }$
C. $\frac { \pi } { 4 }$
D. $\frac { \pi } { 6 }$

In $\triangle A B C$, the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$ respectively. Given $b \sin C + c \sin B = 4 a \sin B \sin C$ and $b ^ { 2 } + c ^ { 2 } - a ^ { 2 } = 8$, then the area of $\triangle A B C$ is \_\_\_\_

In planar quadrilateral $A B C D$, $\angle A D C = 90 ^ { \circ }$, $\angle A = 45 ^ { \circ }$, $A B = 2$, $B D = 5$.
(1) Find $\cos \angle A D B$;
(2) If $D C = 2 \sqrt { 2 }$, find $B C$.

12. In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given $10 \sin A - 5 \sin C = 2 \sqrt { 6 }$ and $\cos B = \frac { 1 } { 5 }$, then $\frac { c } { a } =$
$$\text { A. } \frac { 6 } { 7 } \quad \text{B.} \frac { 7 } { 6 } \quad \text{C.} \frac { 5 } { 6 } \quad \text{D.} \frac { 6 } { 5 }$$

Section II

II. Fill-in-the-Blank Questions: This section contains 4 questions, each worth 5 points, totaling 20 points. Write your answers on the answer sheet.

In $\triangle A B C$, $a = 3 , \quad b - c = 2 , \quad \cos B = - \frac { 1 } { 2 }$. (I) Find the values of $b$ and $c$; (II) Find the value of $\sin ( B - C )$.

15. In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given $b \sin A + a \cos B = 0$, then $B =$ $\_\_\_\_$ .

15. In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. If $b = 6 , a = 2 c , B = \frac { \pi } { 3 }$, then the area of $\triangle A B C$ is $\_\_\_\_$ .

18. (12 points) In $\triangle A B C$ , the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given that $a \sin \frac { A + C } { 2 } = b \sin A$ .
(1) Find $B$ .
(2) If $\triangle A B C$ is an acute triangle and $c = 1$ , find the range of the area of $\triangle A B C$ .

18. (12 points) In $\triangle A B C$ , the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given that When $t = 0$, $S = 3$; when $t = \pm 1$, $S = 4\sqrt{2}$.
Therefore, the area of quadrilateral $ADBE$ is $3$ or $4\sqrt{2}$.

In $\triangle A B C$ , $\cos C = \frac { 2 } { 3 } , A C = 4 , B C = 3$ , then $\cos B =$
A. $\frac { 1 } { 9 }$
B. $\frac { 1 } { 3 }$
C. $\frac { 1 } { 2 }$
D. $\frac { 2 } { 3 }$

In $\triangle A B C$, $\cos C = \frac { 2 } { 3 } , A C = 4 , B C = 3$. Then $\tan B =$
A. $\sqrt { 5 }$
B. $2 \sqrt { 5 }$
C. $4 \sqrt { 5 }$
D. $8 \sqrt { 5 }$

In $\triangle A B C$ , $\sin ^ { 2 } A - \sin ^ { 2 } B - \sin ^ { 2 } C = \sin B \sin C$ .
(1) Find $A$ ;
(2) If $B C = 3$ , find the maximum value of the perimeter of $\triangle A B C$ .

In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given $B = 150 ^ { \circ }$ ,
(1) If $a = \sqrt { 3 } c , b = 2 \sqrt { 7 }$ , find the area of $\triangle A B C$ ;
(2) If $\sin A + \sqrt { 3 } \sin C = \frac { \sqrt { 2 } } { 2 }$ , find $C$ .