8. In $\triangle A B C$, it is known that $B = 120 ^ { \circ } , A C = \sqrt { 19 } , A B = 2$, then $B C =$ ( )
A. 1
B. $\sqrt { 2 }$
C. $\sqrt { 5 }$
D. 3

8. On December 8, 2020, China and Nepal jointly announced that the latest height of Mount Everest is 8848.86 m. The trigonometric height measurement method is one of the methods for measuring the height of Mount Everest. The figure on the right is a schematic diagram of the trigonometric height measurement method. There are three points $A, B, C$, and their projections $A', B', C'$ on the same horizontal plane satisfy $\angle A'C'B' = 45°$, $\angle A'B'C' = 60°$. The angle of elevation from point $C$ to point $B$ is $15°$. The difference between $BB'$ and $CC'$ is 100 m. The angle of elevation from point $B$ to point $A$ is $45°$. Then the height difference between points $A$ and $C$ to the horizontal plane $A'B'C'$ is $AA' - CC'$ approximately equals ($\sqrt{3} \approx 1.732$)
A. 346
B. 373
C. 446
D. 473

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 $a = 2$ .

19.
(1)
By the sine rule:
$$B D = \frac { a \sin C } { \sin B } = \frac { a c } { b } = \frac { b ^ { 2 } } { b } = b$$
(2)
Note that $\angle B D A$ and $\angle B D C$ are supplementary, so their cosines are opposite in sign. Applying the cosine rule in triangles $\triangle B D A$ and $\triangle B D C$:
$$\cos \angle B D A + \cos \angle B D C = \frac { b ^ { 2 } + \left( \frac { 2 } { 3 } b \right) ^ { 2 } - c ^ { 2 } } { 2 \cdot \left( \frac { 2 } { 3 } b \right) \cdot b } + \frac { b ^ { 2 } + \left( \frac { 1 } { 3 } b \right) ^ { 2 } - a ^

In $\triangle A B C$, point $D$ is on side $B C$, $\angle A D B = 120 ^ { \circ } , A D = 2 , C D = 2 B D$ . If $S_{\triangle ABD} : S_{\triangle ACD} = 1 : 2$, then $B D =$ $\_\_\_\_$ .

In $\triangle ABC$, point $D$ is on side $BC$, $\angle ADB = 120 ^ { \circ }$, $AD = 2$, and $CD = 2 BD$. Then $BD =$ $\_\_\_\_$.

(12 points) Let the sides opposite to angles $A, B, C$ of $\triangle ABC$ be $a, b, c$ respectively. Given $$\sin C \sin(A - B) = \sin B \sin(C - A)$$ (1) Prove: $2a^2 = b^2 + c^2$;
(2) If $a = 5, \cos A = \frac{25}{31}$, find the perimeter of $\triangle ABC$.

In the quadrangular pyramid $P - ABCD$ , the base $ABCD$ is a square with $AB = 4$ , $PC = PD = 3$ , $\angle PCA = 45^{\circ}$ , then the area of $\triangle PBC$ is
A. $2\sqrt{2}$
B. $3\sqrt{2}$
C. $4\sqrt{2}$
D. $5\sqrt{2}$

In $\triangle ABC$ , $\angle BAC = 60^{\circ} , AB = 2 , BC = \sqrt{6}$ . $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$ . Then $AD =$ $\_\_\_\_$ .

In $\triangle A B C$, it is given that $\angle B A C = 120 ^ { \circ } , A B = 2 , A C = 1$.
(1) Find $\sin \angle A B C$.
(2) If $D$ is a point on $BC$ such that $\angle B A D = 90 ^ { \circ }$, find the area of $\triangle A D C$.

Let the interior angles $A , B , C$ of $\triangle A B C$ and their opposite sides $a , b , c$ satisfy $\sin A + \sqrt { 3 } \cos A = 2$.
(1) Find $A$.
(2) If $a = 2$ and $\sqrt { 2 } b \sin C = c \sin 2 B$, find the perimeter of $\triangle A B C$.

(13 points) Let the sides opposite to angles $A , B , C$ of $\triangle A B C$ be $a , b , c$ respectively. Given $\sin C = \sqrt { 2 } \cos B , a ^ { 2 } + b ^ { 2 } - c ^ { 2 } = \sqrt { 2 } a b$ .
(1) Find $B$ ;
(2) If the area of $\triangle A B C$ is $3 + \sqrt { 3 }$ , find $c$ .

In $\triangle ABC$, $a = 7$, $A$ is an obtuse angle, $\sin 2B = \frac { \sqrt { 3 } } { 7 } b \cos B$.
(1) Find $\angle A$;
(2) Choose one condition from conditions (1), (2), and (3) below as a given condition and find the area of $\triangle ABC$.
(1) $b = 7$; (2) $\cos B = \frac { 13 } { 14 }$; (3) $c \sin A = \frac { 5 } { 2 } \sqrt { 3 }$. Note: If conditions (1), (2), and (3) are solved separately, only the first solution will be graded.

In $\triangle ABC$, $BC = 2$, $AC = 1 + \sqrt{3}$, $AB = \sqrt{6}$, then $A = $ ( )
A. $45°$
B. $60°$
C. $120°$
D. $135°$

Given that the area of $\triangle ABC$ is $\frac{1}{4}$, if $\cos 2A + \cos 2B + 2\sin C = 2$, $\cos A \cos B \sin C = \frac{1}{4}$, then
A. $\sin C = \sin^2 A + \sin^2 B$
B. $AB = \sqrt{2}$
C. $\sin A + \sin B = \frac{\sqrt{6}}{2}$
D. $AC^2 + BC^2 = 3$

104- What is the area of a triangle with sides $7, 9, 12$?
(1) $15\sqrt{2}$ (2) $14\sqrt{3}$ (3) $12\sqrt{5}$ (4) $14\sqrt{5}$

129. In an isosceles triangle, if the altitude is 8 units and the radius of the inscribed circle is 3 units, what is the length of the base of this triangle?

• [(1)] $15$
• [(2)] $12$
• [(3)] $14$
• [(4)] $16$

128. In triangle $ABC$, the side lengths are $BC = 9$, $AC = 8$, and $AB = 2$. The angle bisectors of angle $A$ intersect side $BC$ at $M$ and $N$. What is the length of $MN$?
(1) $4/2$ (2) $4/5$ (3) $4/8$ (4) $5/1$

137- The internal bisector of angle $A$ in triangle $ABC$ divides the opposite side into segments of $3/5$ and $2/5$ units. If the measure of angle $C$ is $60$ degrees, the smaller side of the triangle is how many units?
\[ \text{(1)}\ 3/75 \qquad \text{(2)}\ 4/25 \qquad \text{(3)}\ 4/75 \qquad \text{(4)}\ 5/25 \]

11. Triangle $ABC$ has sides $\sqrt{3}$, $6$, and $\alpha$ (with the angle between them being $\alpha$) and is inscribable in a circle. If the area of this triangle is at most $4.5$, what is the minimum value of $\alpha$?
(1) $2$ (2) $3$ (3) $4$ (4) $5$
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25. In a rhombus, the geometric mean of the two diagonals is the length of the rhombus's side. The smaller angle of the triangle formed by drawing the diagonals of this rhombus is how many degrees?
(1) $10$ (2) $15$ (3) $30$ (4) $45$

28. In triangle $ABC$, the medians drawn from vertices $B$ and $C$ are perpendicular to each other. If the length of the median drawn from vertex $C$ is $4.5$ and the area of this triangle is $18$, what is the ratio of the lengths of the medians drawn from vertices $B$ and $C$?
(1) $\dfrac{17}{9}$ (2) $\dfrac{19}{9}$ (3) $\dfrac{5}{3}$ (4) $\dfrac{4}{3}$

31. Among triangles with area 35 square units that have a side of length 15 units in common, what is the minimum perimeter?

• [(1)] $30$
• [(2)] $32$
• [(3)] $34$
• [(4)] $36$

32. In the figure below, if $\widehat{DAC} = 3\widehat{BAD}$, what is the length of side $AC$?
[Figure: Triangle with vertices $A$, $B$, $C$, point $D$ on $BC$ with $BD = 4$, $AB = 8$, $AD = 6$]

• [(1)] $19/2$
• [(2)] $16/8$
• [(3)] $18/6$
• [(4)] $15/4$

Let $A_1, A_2, \ldots, A_n$ be the vertices of a regular polygon and $A_1A_2$, $A_2A_3$, $\ldots$, $A_{n-1}A_n$, $A_nA_1$ be its $n$ sides. If $\left(\frac{1}{A_1A_2}\right) - \left(\frac{1}{A_1A_4}\right) = \frac{1}{A_1A_3}$, then the value of $n$ is
(a) 5
(b) 6
(c) 7
(d) 8