<選填題>等腰三角形\(ABC\)中，令\(\theta=\angle BAC\)。若\(\overline{AB}^{2}=\overline{AC}^{2}=\overline{BC}=\sin\theta\)，（求三角形面積，答案部分原表述不清，按照正確思路解題）
答案
$\dfrac{8}{25}$
$\cos \theta = {\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\times \overline{AB}\times \overline{AC}} ={\sin \theta +\sin \theta-\sin^2\theta \over 2\times \sqrt{\sin \theta}\times \sqrt{\sin \theta}} ={2\sin\theta -\sin^2\theta \over 2\sin \theta} =1-{1\over 2}\sin \theta\\ 又\cos^2\theta +\sin^2\theta =1 \Rightarrow (1-{1\over 2}\sin \theta)^2 +\sin^2\theta =1 \Rightarrow {5\over 4}\sin^2\theta-\sin \theta=0 \\\Rightarrow \sin\theta({5\over 4}\sin\theta-1)=0 \Rightarrow \sin \theta= {4\over 5} \Rightarrow \triangle ABC 面積= {1\over 2}\times \overline{AB}\times \overline{AC} \sin \theta \\ ={1\over 2}\sin^2 \theta ={1\over 2}\times {16\over 25} =\bbox[red, 2pt]{8\over 25}$
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