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114-學測數學模考_北模_17


<選填題>設 \(\triangle ABC\) 中,\(\angle A\)、\(\angle B\)、\(\angle C\) 的對邊邊長分別為 \(a\)、\(b\)、\(c\),已知 \(b^2 = ac\),若點 \(D\) 在 \(\overline{AC}\) 邊上,且 \(\overline{BD} = \overline{AC}\),\(\overline{AD} = 3\overline{CD}\),試求 $\cos\angle ABC = \frac{~~~~~~~~~~~}{~~~~~~~~~~}$(化為最簡分數)

答案

設 \(CD = x\),則 \(AD = 3x\),\(AC = b = 4x\),\(BD = b = 4x\)。在 \(\triangle BCD\) 與 \(\triangle ABC\) 用餘弦定理,聯立 \(b^2 = ac\),得 \(12a^2 - 19ac + 4c^2 = 0\),解得 \(a = \frac{4c}{3}\)。代入 \(\cos\angle ABC = \frac{a^2 + c^2 - b^2}{2ac}\),得 \(\frac{13}{24}\)。答案:\(\frac{13}{24}\) 報錯
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