Косинусова теорема задача

[Гост]

Дадено: ABCD успоредник, BC е 4 см, АС е 4√7, BD=4√3. Намерете периметъра и ъгъл ABC.

[ammornil]

Дадено: ABCD успоредник, BC е 4 см, АС е 4√7, BD=4√3. Намерете периметъра и ъгъл ABC.

$\\[12pt] AC= 4\sqrt{7}[cm], \quad BD=4\sqrt{3}[cm], \quad AB= CD= x[cm], \quad BC= AD= 4[cm], \quad \angle{ABC}=\alpha \\[6pt] \cos{(180^{\circ}-\angle{ABC})}= -\cos{(\angle{ABC})}$ $$ \begin{array}{|l} AC^{2}= AB^{2} +BC^{2} -2\cdot{}AB\cdot{}BC\cdot{}\cos{(180^{\circ}-\angle{ABC})} \\[6pt] BD^{2}= AB^{2} +AD^{2} -2\cdot{}AB\cdot{}AD\cdot{}\cos{(\angle{ABC})} \end{array} $$ $\\[12pt] \begin{array}{|l} 112= x^{2} +4^{2} -2\cdot{}x\cdot{}4\cdot{}(-\cos{(\alpha)}) \\[6pt] 48= x^{2} +4^{2} -2\cdot{}x\cdot{}4\cdot{}\cos{(\alpha)} \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} 112= x^{2} +16 +8x\cos{(\alpha)} \\[6pt] 48= x^{2} +16 -8x\cos{(\alpha)} \end{array} \large\} +\normalsize \quad \Leftrightarrow \quad \begin{array}{|l} 160= 2x^{2} +32 \\[6pt] \cos{(\alpha)}=\dfrac{x^{2}-32}{8x} \end{array} \\[6pt] \quad \Leftrightarrow \quad \begin{array}{|l} x^{2}=64 \\[6pt] \cos{(\alpha)}=\dfrac{x^{2}-32}{8x} \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} x=8 \\[6pt] \cos{(\alpha)}=\dfrac{1}{2} \end{array} \Rightarrow \\[6pt] AB= CD= 8[cm] \\[6pt] \angle{ABC}= \alpha= \arccos{\left(\dfrac{1}{2}\right)}= 60^{\circ}$ $$ P_{ABCD}=2\cdot{}AB +2\cdot{}BC= 24[cm], \quad \angle{ABC}= 60^{\circ} $$

[ammornil]

В горното има логическа грешка. Срещу диагонала $AC$ лежи $\angle{ABC}$, а не $180^{\circ}-\angle{ABC}$. Оттам вярната система е:$$ \begin{array}{|l} AC^{2}= AB^{2} +BC^{2} -2\cdot{}AB\cdot{}BC\cdot{}\cos{(\angle{ABC})} \\[6pt] BD^{2}= AB^{2} +AD^{2} -2\cdot{}AB\cdot{}AD\cdot{}\cos{(180^{\circ}-\angle{ABC})} \end{array} $$ $\\[12pt] \begin{array}{|l} 112= x^{2} +4^{2} -2\cdot{}x\cdot{}4\cdot{}\cos{(\alpha)} \\[6pt] 48= x^{2} +4^{2} -2\cdot{}x\cdot{}4\cdot{}\cos{(180^{\circ}-\alpha)} \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} 112= x^{2} +16 -8x\cos{(\alpha)} \\[6pt] 48= x^{2} +16 +8x\cos{(\alpha)} \end{array} \large\} +\normalsize \quad \Leftrightarrow \quad \begin{array}{|l} 160= 2x^{2} +32 \\[6pt] \cos{(\alpha)}=-\dfrac{x^{2}-32}{8x} \end{array} \\[6pt] \quad \Leftrightarrow \quad \begin{array}{|l} x^{2}=64 \\[6pt] \cos{(\alpha)}=-\dfrac{x^{2}-32}{8x} \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} x=8 \\[6pt] \cos{(\alpha)}=-\dfrac{1}{2} \end{array} \Rightarrow \\[6pt] AB= CD= 8[cm] \\[6pt] \angle{ABC}= \alpha= \arccos{\left(-\dfrac{1}{2}\right)}= 120^{\circ} $ $$ P_{ABCD}=2\cdot{}AB +2\cdot{}BC= 24[cm], \quad \angle{ABC}= 120^{\circ} $$