题目

Problem

A parallelogram $ABCD$ has area $40\text{ cm}^2$ .

Given that $AB$ has length $10$ cm, $BC$ has length $6$ cm and angle $DAB$ is obtuse, find

(a) the size of angle $DAB$ , in degrees, to 2 decimal places,

(3)

(b) the length of diagonal $BD$ , in cm, to one decimal place.

(2)

解答

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平行四边形面积可以写成 $ab\sin\theta$ 。这里相邻边是 $10$ 和 $6$ ，夹角是 $\angle DAB$ 。因为题目说它是钝角，所以要取钝角解。

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Let $\theta=\angle DAB$ . Then

\begin{align*} 40=&\,10(6)\sin\theta\\ \sin\theta=&\,\frac23. \end{align*}

The acute solution is

\begin{align*} \sin^{-1}\left(\frac23\right)=41.810\ldots^\circ. \end{align*}

Since $\theta$ is obtuse,

\begin{align*} \theta =&\,180^\circ-41.810\ldots^\circ\\ =&\,138.19^\circ. \end{align*}

展开

在三角形 $ABD$ 中， $AB=10$ ， $AD=BC=6$ ，夹角就是刚求出的 $\angle DAB$ 。用余弦定理求 $BD$ 。

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By the cosine rule,

\begin{align*} BD^2 =&\,10^2+6^2-2(10)(6)\cos(138.19^\circ)\\ =&\,225.44\ldots. \end{align*}

Therefore

\begin{align*} BD=15.014\ldots=15.0\text{ cm}. \end{align*}