Posted in

107學測數學考科-10


<多選題>已知坐標平面上 \(\triangle ABC\),其中 \(\overset{\rightharpoonup}{AB} = (-4,3)\),且 \(\overset{\rightharpoonup}{AC} = \left( \frac{2}{5}, \frac{4}{5} \right)\)。試選出正確的選項。
(1) \(\overline{BC} = 5\)
(2) \(\triangle ABC\) 是直角三角形
(3) \(\triangle ABC\) 的面積為 \(\frac{11}{5}\)
(4) \(\sin B \gt \sin C\)
(5) \(\cos A \gt \cos B\)。

答案

(1) \( \overset{\rightharpoonup}{BC} = \overset{\rightharpoonup}{AC} - \overset{\rightharpoonup}{AB} = \left( \frac{22}{5}, -\frac{11}{5} \right) \),長度 \( \frac{11\sqrt{5}}{5} \neq 5 \) ✗。
(2) \( \overline{AB}=5 \),\( \overline{AC}=\frac{2\sqrt{5}}{5} \),\( \overline{BC}=\frac{11\sqrt{5}}{5} \),檢查得 \( \overline{AC}^2 + \overline{BC}^2 = \frac{4}{5} + \frac{121}{5} = 25 = \overline{AB}^2 \) ✓。
(3) 面積 \( \frac{1}{2} \times \frac{2\sqrt{5}}{5} \times \frac{11\sqrt{5}}{5} = \frac{11}{5} \) ✓。
(4) \( \sin B = \frac{\overline{AC}}{\overline{AB}} = \frac{2\sqrt{5}}{25} \),\( \sin C = 1 \),故 \( \sin B \lt \sin C \) ✗。
(5) \( \cos A = \frac{\overset{\rightharpoonup}{AB} \cdot \overset{\rightharpoonup}{AC}}{5 \cdot \frac{2\sqrt{5}}{5}} = \frac{-8/5+12/5}{2\sqrt{5}} = \frac{4/5}{2\sqrt{5}} = \frac{2}{5\sqrt{5}} \),\( \cos B = \frac{\overset{\rightharpoonup}{BA} \cdot \overset{\rightharpoonup}{BC}}{5 \cdot \frac{11\sqrt{5}}{5}} = \frac{ (4,-3) \cdot (22/5,-11/5) }{11\sqrt{5}} = \frac{88/5+33/5}{11\sqrt{5}} = \frac{121/5}{11\sqrt{5}} = \frac{11}{5\sqrt{5}} \),故 \( \cos A \lt \cos B \) ✗。
故選(2)(3)。答案:(2)(3) 報錯
ChatGPT    DeepSeek


我要來個錯題通知
Powered by