AMC12 2010 B

AMC12 2010 B · Q13

AMC12 2010 B · Q13. It mainly tests Triangles (properties), Trigonometry (basic).

In $\triangle ABC$, $\cos(2A-B)+\sin(A+B)=2$ and $AB=4$. What is $BC$?

在 $\triangle ABC$ 中，$\cos(2A-B)+\sin(A+B)=2$ 且 $AB=4$。求 $BC$？

(A) $\sqrt{2}$ $\sqrt{2}$

(B) $\sqrt{3}$ $\sqrt{3}$

(D) $2\sqrt{2}$ $2\sqrt{2}$

(E) $2\sqrt{3}$ $2\sqrt{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

We note that $-1$ $\le$ $\sin x$ $\le$ $1$ and $-1$ $\le$ $\cos x$ $\le$ $1$. Therefore, there is no other way to satisfy this equation other than making both $\cos(2A-B)=1$ and $\sin(A+B)=1$, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easily conclude that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. It is clear that $\triangle ABC$ is a $30^{\circ},60^{\circ},90^{\circ}$ triangle with $BC=2$ $\Longrightarrow$ $(C)$.

注意到 $-1$ $\le$ $\sin x$ $\le$ $1$ 且 $-1$ $\le$ $\cos x$ $\le$ $1$。因此，满足该方程的唯一方式是同时令 $\cos(2A-B)=1$ 且 $\sin(A+B)=1$，因为任何其他方式都会使其中一个值大于 1，这与前述结论矛盾。由此可得 $2A-B=0^{\circ}$ 且 $A+B=90^{\circ}$，解该方程组得到 $A=30^{\circ}$ 且 $B=60^{\circ}$。显然 $\triangle ABC$ 是一个 $30^{\circ},60^{\circ},90^{\circ}$ 三角形，且 $BC=2$ $\Longrightarrow$ $(C)$。

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