高専の数学解説

第11章 11.26

(1)

∠BAC=arg⁡γ−αβ−α=arg⁡(4−7i)−(2+3i)(5+4i)−(2+3i)=arg⁡2+4i3+i=arg⁡(2+4i)(3−i)10=arg⁡(1+i)\angle BAC = \arg \frac{\gamma - \alpha}{\beta - \alpha} = \arg \frac{(4-7i)-(2+3i)}{(5+4i)-(2+3i)} = \arg \frac{2+4i}{3+i} = \arg \frac{(2+4i)(3-i)}{10} = \arg(1+i)

よって ∠BAC=π4\angle BAC = \frac{\pi}{4}

(2)

∠BAC=arg⁡γ−αβ−α=arg⁡(3+i)−(−2i)(3−i)−(−2i)=arg⁡3+3i3+i=arg⁡(3+3i)(3−i)4=arg⁡12(3+3i)\angle BAC = \arg \frac{\gamma - \alpha}{\beta - \alpha} = \arg \frac{(\sqrt{3}+i)-(-2i)}{(\sqrt{3}-i)-(-2i)} = \arg \frac{\sqrt{3}+3i}{\sqrt{3}+i} = \arg \frac{(\sqrt{3}+3i)(\sqrt{3}-i)}{4} = \arg \frac{1}{2}(3+\sqrt{3}i)

=arg⁡3(32+12i)= \arg \sqrt{3}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)

よって ∠BAC=π6\angle BAC = \frac{\pi}{6}

(3)

∠BAC=arg⁡γ−αβ−α=arg⁡(2−5i)−(5−4i)(6−2i)−(5−4i)=arg⁡−3−i1+2i=arg⁡(−3−i)(1−2i)5=arg⁡(−1+i)\angle BAC = \arg \frac{\gamma - \alpha}{\beta - \alpha} = \arg \frac{(2-5i)-(5-4i)}{(6-2i)-(5-4i)} = \arg \frac{-3-i}{1+2i} = \arg \frac{(-3-i)(1-2i)}{5} = \arg(-1+i)

よって ∠BAC=34π\angle BAC = \frac{3}{4}\pi

解説: r31bn1z