AMC12 2012 A

AMC12 2012 A · Q10

AMC12 2012 A · Q10. It mainly tests Triangles (properties), Area & perimeter.

A triangle has area $30$, one side of length $10$, and the median to that side of length $9$. Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$?

一个三角形的面积为 $30$，有一边长为 $10$，且到该边的中线长为 $9$。设 $\theta$ 为该边与中线所成的锐角。求 $\sin{\theta}$。

(A) \frac{3}{10} \frac{3}{10}

(B) \frac{1}{3} \frac{1}{3}

(D) \frac{2}{3} \frac{2}{3}

(E) \frac{9}{10} \frac{9}{10}

Answer

Correct choice: (D)

正确答案：（D）

Solution

$AB$ is the side of length $10$, and $CD$ is the median of length $9$. The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$. To find $\sin{\theta}$, just use opposite over hypotenuse with the right triangle $\triangle DCE$. This is equal to $\frac69=\boxed{\textbf{(D)}\ \frac23}$.

设 $AB$ 为长度为 $10$ 的边，$CD$ 为长度为 $9$ 的中线。因为 $\frac12(底)(高)=30$，所以从 $C$ 到 $AB$ 的高为 $6$。$\theta$ 为 $\angle CDE$。在直角三角形 $\triangle DCE$ 中，$\sin{\theta}$ 等于对边比斜边，即 $\frac69=\boxed{\textbf{(D)}\ \frac23}$。

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