【数学I】例題4.1.10：2直線のなす角（One More）★★

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% 例題I4.1.10：2直線のなす角（One More）★★
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2直線$y=\sqrt{3}x+2\cdots(\mathrm{i}),y=-x+1\cdots(\mathrm{ii})$について，次の問いに答えよ．

(1) 直線(i)が$x$軸の正の向きとのなす角を求めよ．

(2) 2直線(i)，(ii)のなす角$\theta(0^{\circ}\leqq\theta\leqq 90^{\circ})$を求めよ．

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(1) 直線(i)が$x$軸の正の向きとのなす角を$\alpha$とすると，$\tan\alpha=\sqrt{3}$

$0^{\circ}\leqq\alpha<180^{\circ}$より，$\alpha=60^{\circ}$

(2) 直線(ii)が$x$軸の正の向きとのなす角を$\beta$とすると，$\tan\beta=-1$

$0^{\circ}\leqq\beta<180^{\circ}$より，$\beta={135^{\circ}}$

右の図から，求める角$\theta$は，$0^{\circ}\leqq\theta\leqq 90^{\circ}$より，

\[
\theta=\beta-\alpha=135^\circ-60^\circ=75^\circ
\]

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% 問題I4.1.10：2直線のなす角（One More）★★
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2直線$x-\sqrt{3}y=0\cdots(\mathrm{i}),x+\sqrt{3}y=0\cdots(\mathrm{ii})$について，次の問いに答えよ．

(1) 直線(i)が$x$軸の正の向きとのなす角を求めよ．

(2) 2直線(i)，(ii)のなす角$\theta(0^{\circ}\leqq\theta\leqq 90^{\circ})$を求めよ．

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(1) 直線(i)が$x$軸の正の向きとのなす角を$\alpha$とすると，$y=\frac{1}{\sqrt{3}}x$より，$\tan\alpha=\frac{1}{\sqrt{3}}$

$0^{\circ}\leqq\alpha<180^{\circ}$より，$\alpha=30^{\circ}$

(2) 直線(ii)が$x$軸の正の向きとのなす角を$\beta$とすると，$y=-\frac{1}{\sqrt{3}}x$より，$\tan\beta=-\frac{1}{\sqrt{3}}$

$0^{\circ}\leqq\beta<180^{\circ}$より，$\beta={150^{\circ}}$

右の図から，$\beta-\alpha=150^\circ-30^\circ={120^\circ}$

よって，求める角$\theta$は，$0^{\circ}\leqq\theta\leqq 90^{\circ}$より，$\theta=180^\circ-120^\circ=60^\circ$

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% 例題I4.1.10の側注：2直線のなす角（One More）★★
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% 側注1
$y=\sqrt{3}x+2$の傾きは，$y=\sqrt{3}x$の傾きと同じで，$\sqrt{3}$である．

% 側注2
$y=-x+1$の傾きは，$y=-x$の傾きと同じで，$-1$である．

% 側注3
求める角は，2直線の図をかいて判断するとよい．

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% 側注1
求める角は，2直線の図をかいて判断するとよい．

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% 例題I4.1.10の考え方・補足など：2直線のなす角（One More）★★
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直線$y=mx$と$x$軸の正の向きとのなす角が$\theta(0^{\circ}\leqq\theta<180^{\circ},\theta\neq 90^{\circ})$のとき，$m=\tan\theta$であることを利用できる．また，直線は平行移動しても傾きが変わらないことから，「直線$y=mx+n$と$x$軸の正の向きとのなす角」は，「直線$y=mx$と$x$軸の正の向きとのなす角」を考えればよい．

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2直線のなす角$\theta$について，傾きである$\tan\theta$を考える．

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とくに断りがない場合は，2直線のなす角$\theta$は，$0^{\circ}\leqq\theta\leqq 90^{\circ}$の範囲内で答えることになっているので注意すること（例えば，(2)では丁寧に「なす角$\theta(0^{\circ}\leqq\theta\leqq 90^{\circ})$を求めよ」と断りがあった）．また，「直線と$x$軸とのなす角」は，直線の$y\geqq 0$の部分と$x$軸の正の向きとのなす角のことである．