题目

Problem

The sides of a triangle are in the ratio $2:4:5$ .

(a) Find the size of the largest angle of the triangle, in degrees, to one decimal place.

(2)

Given that the area of the triangle is $140$ cm $^2$

(b) find the length of the shortest side, in cm, to one decimal place.

(3)

解答

(a)

解法一

思路

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最大角一定对着最长边。设三边为 $2,4,5$ 的同倍数即可；因为比例相同，角度不受倍数影响。用余弦法则直接求最长边 $5$ 对应的角。

答题过程

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Let the largest angle be $\theta$ . It is opposite the side in ratio $5$ .

Using the cosine rule,

\begin{align*} 5^2=&\,2^2+4^2-2(2)(4)\cos\theta\\ 25=&\,4+16-16\cos\theta\\ 16\cos\theta=&\,-5\\ \cos\theta=&\,-\frac{5}{16}. \end{align*}

Hence

\begin{align*} \theta=108.2^\circ \end{align*}

to one decimal place.

解法二

思路

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也可以先求一个较小角，再用正弦法则求最大角。这个方法计算多一点，但能训练“最长边对最大角”的对应关系。

答题过程

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Let $A$ be the angle opposite the side in ratio $2$ .

Using the cosine rule,

\begin{align*} 2^2=&\,4^2+5^2-2(4)(5)\cos A\\ 4=&\,16+25-40\cos A\\ 40\cos A=&\,37\\ A=&\,22.33\ldots^\circ. \end{align*}

Let $\theta$ be the largest angle, opposite the side in ratio $5$ .

Using the sine rule,

\begin{align*} \frac{2}{\sin A} =&\,\frac{5}{\sin\theta}\\ \sin\theta =&\,\frac{5\sin(22.33\ldots^\circ)}{2}. \end{align*}

This gives the obtuse angle

\begin{align*} \theta=108.2^\circ \end{align*}

to one decimal place.

(b)

解法一

思路

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设三边为 $2x,4x,5x$ 。用两边夹角面积公式 $\frac12ab\sin C$ ，其中 $2x$ 和 $4x$ 夹着最大角。

答题过程

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Let the side lengths be

\begin{align*} 2x,\quad4x,\quad5x. \end{align*}

Using the angle from part (a),

\begin{align*} 140=&\,\frac12(2x)(4x)\sin(108.2\ldots^\circ)\\ 140=&\,4x^2\sin(108.2\ldots^\circ). \end{align*}

\begin{align*} x^2 =&\,\frac{140}{4\sin(108.2\ldots^\circ)}\\ =&\,36.845\ldots \end{align*}

and

\begin{align*} x=6.070\ldots \end{align*}

The shortest side is $2x$ , so

\begin{align*} 2x=&\,12.140\ldots\\ =&\,12.1\text{ cm} \end{align*}

to one decimal place.

IAL 2025 Oct Q3

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程