题目

Problem

In triangle $ABC$ ,

$AB=15$ cm, $AC=25$ cm and angle $BAC=\theta^\circ$ .

Given that the area of triangle $ABC$ is $100$ cm $^2$ ,

(a) find the value of $\sin\theta^\circ$ .

(2)

Given that $\theta>90$ ,

(b) find the length of $BC$ , giving your answer to 3 significant figures.

(3)

解答

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已知两边和夹角，三角形面积用 $\frac12 ab\sin C$ 。这里夹角正好是 $AB$ 与 $AC$ 之间的 $\theta^\circ$ 。

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Using the area formula,

\begin{align*} \text{Area} =&\,\frac12(AB)(AC)\sin\theta^\circ. \end{align*}

Substitute the given values:

\begin{align*} 100=&\,\frac12(15)(25)\sin\theta^\circ\\ 100=&\,\frac{375}{2}\sin\theta^\circ\\ \sin\theta^\circ=&\,\frac{200}{375}\\ =&\,\frac{8}{15}. \end{align*}

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求第三边 $BC$ ，要用余弦定理。由于 $\theta>90^\circ$ ，所以 $\cos\theta^\circ$ 是负数；已知 $\sin\theta^\circ=\frac8{15}$ 后，要选负的余弦值。

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Since $\theta>90^\circ$ ,

\begin{align*} \cos\theta^\circ =&\,-\sqrt{1-\sin^2\theta^\circ}\\ =&\,-\sqrt{1-\left(\frac{8}{15}\right)^2}\\ =&\,-\sqrt{\frac{161}{225}}\\ =&\,-\frac{\sqrt{161}}{15}. \end{align*}

Using the cosine rule,

\begin{align*} BC^2 =&\,15^2+25^2-2(15)(25)\cos\theta^\circ\\ =&\,225+625-750\left(-\frac{\sqrt{161}}{15}\right)\\ =&\,850+50\sqrt{161}. \end{align*}

Therefore

\begin{align*} BC=&\,\sqrt{850+50\sqrt{161}}\\ =&\,38.5\text{ cm}\quad\text{to 3 significant figures.} \end{align*}