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 $BD =$ $\_\_\_\_$.

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

(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. $AC^2 + BC^2 = 3$

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. $AC^2 + BC^2 = 3$

In a right angle triangle with sides $a < b < c$, where $\angle ACB = \theta$ is the smallest angle, show that $\sin^2\theta - \sqrt{5}\sin\theta + 1 = 0$, given that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ (i.e., the reciprocals of the sides also form a right triangle).

In a triangle with angles $P$, $Q$, $R$, let $\alpha$, $\beta$, $\gamma$ be the angles $\angle QCR = 2P$, $\angle QIR = Q + R$, $\angle QOR = P + Q/2 + R/2$ respectively. Show that $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} > \frac{1}{45}$.

In a triangle, $E$ is the midpoint of $AC$. Let $\angle BCE = \angle ABE$. Prove that $AB + BD = CD$ (where $D$ is the midpoint of $BC$), i.e., $AB + BD = l_1 + l_2$.

Three triangles are formed by drawing lines from the vertices of a triangle $ABC$ to the opposite sides, each making equal angles with the sides. Let $\Delta_1, \Delta_2, \Delta_3$ be the areas of the three smaller triangles formed at the vertices with the circumradius equal to 1.
a) Express the total area $\Delta = \Delta_1 + \Delta_2 + \Delta_3$ in terms of $A, B, C$.
b) Find the angles $A, B, C$ that maximize $\Delta$.
c) Verify that the maximum occurs for an isosceles triangle and prove $C = A$ by calculus.

Let $ABC$ be a triangle $A \neq B$ and let $P \in (AB)$ be a point for which denote $m(\widehat{ACP}) = x$ and $m(\widehat{BCP}) = y$. Prove that $\frac{\sin A \sin B}{\sin(A-B)} = \frac{\sin x \sin y}{\sin(x-y)}$ if and only if $PA = PB$.

An aeroplane $P$ is moving in the air along a straight line path which passes through the points $P_1$ and $P_2$, and makes an angle $\alpha$ with the ground. Let $O$ be the position of an observer. When the plane is at the position $P_1$ its angle of elevation is $30^\circ$ and when it is at $P_2$ its angle of elevation is $60^\circ$ from the position of the observer. Moreover, the distances of the observer from the points $P_1$ and $P_2$ respectively are 100 metres and $500/3$ metres.
Then $\alpha$ is equal to
(a) $\tan^{-1}\{(2-\sqrt{3})/(2\sqrt{3}-1)\}$
(b) $\tan^{-1}\{(2\sqrt{3}-3)/(4-2\sqrt{3})\}$
(c) $\tan^{-1}\{(2\sqrt{3}-2)/(5-\sqrt{3})\}$
(d) $\tan^{-1}\{(6-\sqrt{3})/(6\sqrt{3}-1)\}$

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

The sides of a triangle are given to be $x ^ { 2 } + x + 1, 2x + 1$ and $x ^ { 2 } - 1$. Then the largest of the three angles of the triangle is
(A) $75 ^ { \circ }$
(B) $\left( \frac { x } { x + 1 } \pi \right)$ radians
(C) $120 ^ { \circ }$
(D) $135 ^ { \circ }$

Two poles, $AB$ of length two metres and $CD$ of length twenty metres are erected vertically with bases at $B$ and $D$. The two poles are at a distance not less than twenty metres. It is observed that $\tan \angle ACB = 2/77$. The distance between the two poles is
(A) $72 m$
(B) 68 m
(C) 24 m
(D) 24.27 m