1959 AMC 12 第 36 题

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36.

一个三角形的底边长为 8080,其中一个底角为 6060^\circ,另两边长度之和为 9090。最短边长为:

The base of a triangle is 80,80, and one of the base angles is 60.60^\circ. The sum of the lengths of the other two sides is 90.90. The shortest side is:

4545

4040

3636

1717

1212

答案:D
知识点:余弦定理代数变形
难度评级:1550
小提示:

设与 6060^\circ 角相邻的边长为 xx,则第三边长为 90x90-x

Let the side adjacent to the 6060^\circ angle be xx, so the third side is 90x90-x

大提示:

对夹角两边 8080xx 应用余弦定理

Apply the law of cosines with included sides 8080 and xx

解答:

设与 6060^\circ 底角相邻的边长为 xx,对边长为 90x90-x。由余弦定理 (90x)2=802+x22(80)(x)cos60 \begin{aligned} (90-x)^2 &=80^2+x^2\\ &\quad{}-2(80)(x)\cos60^\circ \end{aligned}\text{。}化简得 8100180x=640080x8100-180x=6400-80x,所以 x=17x=17。三边长为 17, 7317,\ 738080,最短边为 1717

因此,正确答案是 D

Let the side adjacent to the 6060^\circ base angle be x,x, and let the opposite side be 90x.90-x. The law of cosines gives (90x)2=802+x22(80)(x)cos60. \begin{aligned} (90-x)^2 &=80^2+x^2\\ &\quad{}-2(80)(x)\cos60^\circ. \end{aligned} Simplifying yields 8100180x=640080x,8100-180x=6400-80x, so x=17.x=17. The side lengths are 17, 73,17,\ 73, and 80,80, and the shortest is 17.17.

Thus, the correct answer is D.

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