In $\triangle A B C$, let the sides opposite to angles $\mathrm { A } , \mathrm { B } , \mathrm { C }$ be $a , b , c$ respectively. Given $a = 2 , \cos C = - \frac { 1 } { 4 } , 3 \sin A = 2 \sin B$, then $\mathrm { c } =$ $\_\_\_\_$ .
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 $\cos \left( 2 A + \frac { \pi } { 6 } \right)$.
In $\triangle \mathrm{ABC}$, the angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and their opposite sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ respectively. Given that the area of $\triangle \mathrm{ABC}$ is $3\sqrt{15}$, $b - c = 2$, $\cos A = -\frac{1}{4}$, then the value of $a$ is .
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 $\triangle A B C$, $a = 3 , \quad b - c = 2 , \quad \cos B = - \frac { 1 } { 2 }$. (I) Find the values of $b$ and $c$; (II) Find the value of $\sin ( B - C )$.
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
Let the quadrangle ABCD be a rhombus where the length of the sides is $\sqrt { 2 }$ and $\angle \mathrm { ABC } = 30 ^ { \circ }$. (1) We have $$\mathrm { AC } ^ { 2 } = \mathbf { A } - \mathbf { B } \sqrt { \mathbf { C } } , \quad \mathrm { BD } ^ { 2 } = \mathbf { E } + \mathbf{F} \sqrt{\mathbf{E}} .$$ Now, for any positive numbers $a$ and $b$, we have $$( \sqrt { a } \pm \sqrt { b } ) ^ { 2 } = a + b \pm 2 \sqrt { a b } \quad \text { (double-sign correspondence). }$$ Using this formula, we obtain $$\mathrm { AC } = \sqrt { \mathbf { G } } - \mathbf { H } , \quad \mathrm { BD } = \sqrt { \mathbf { I } } + \mathbf{I} . \mathbf { J } .$$ (2) Let us draw four circles, each centered on one vertex of rhombus ABCD, with the following conditions: The radii of the circles centered on vertices A and C are of length $r$, and those centered on vertices B and D are of length $\sqrt { 2 } - r$. Circles centered on opposite vertices (A and C, B and D) may touch each other but may not intersect. Let us denote the area of the region common to rhombus ABCD and these four circles by $S$. We have $$S = \pi \left( r ^ { 2 } - \frac { \sqrt { \mathbf { K } } } { \mathbf { L } } r \right)$$ where the range of $r$ is $$\sqrt { \mathbf { O } } - \frac { \sqrt { \mathbf { P } } + \mathbf { Q } } { \mathbf { R } } \leqq r \leqq \frac { \sqrt { \mathbf { S } } - \square \mathbf { T } } { \square \mathbf { U } }$$ Hence $S$ is minimized when $r = \frac { \sqrt { \mathbf { V } } } { \mathbf { W } }$, and the value of $S$ then is $\frac { \mathbf{X} } { \mathbf { Y } } \pi$.
In a triangle, one interior angle measure equals the average of the measures of the other two interior angles. The shortest and longest sides of this triangle are 10 and 16 units long, respectively. Accordingly, what is the length of the third side of this triangle in units? A) 11 B) 12 C) 13 D) 14 E) 15