【数学I】例題4.1.1：直角三角形の三角比（One More）★

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% 例題I4.1.1：直角三角形の三角比（One More）★
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(1) 右の図のような三角形において，$\sin A,\cos A,\tan A$の値を求めよ．

(i)

(ii)

(2) 右の図のような三角形において，$x,y$の値を求めよ．

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(1)

(i) 三平方の定理より，$\mathrm{AB}=\sqrt{(\sqrt{5})^2+2^2}=3$

よって，
\[
\sin A=\frac{2}{3},\cos A=\frac{\sqrt{5}}{3},\tan A=\frac{2}{\sqrt{5}}
\]

(ii) 三平方の定理より，$\mathrm{BC}=\sqrt{13^2-12^2}=5$

よって，
\[
\sin A=\frac{5}{13},\cos A=\frac{12}{13},\tan A=\frac{5}{12}
\]

(2) $\sin 30^{\circ}=\frac{x}{6}$であるから，$x=6\sin 30^{\circ}=6\cdot\frac{1}{2}=3$

$\cos 30^{\circ}=\frac{y}{6}$であるから，$y=6\cos 30^{\circ}=6\cdot\frac{\sqrt{3}}{2}=3\sqrt{3}$

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% 問題I4.1.1：直角三角形の三角比（One More）★
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(1) 右の図のような三角形において，$\sin A,\cos A,\tan A$の値を求めよ．

(2) 右の図のような三角形において，$x,y$の値を求めよ．

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% 問題I4.1.1の解答（One More）★
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(1) 三平方の定理より，$\mathrm{AC}=\sqrt{5^2-4^2}=3$

よって，$\sin A=\frac{4}{5},\cos A=\frac{3}{5},\tan A=\frac{4}{3}$

(2) $\sin 60^{\circ}=\frac{x}{10}$であるから，$x=10\sin 60^{\circ}=10\cdot\frac{\sqrt{3}}{2}=5\sqrt{3}$

$\cos 60^{\circ}=\frac{y}{10}$であるから，$y=10\cos 60^{\circ}=10\cdot\frac{1}{2}=5$

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% 例題I4.1.1の側注：直角三角形の三角比（One More）★
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% 側注1
$\tan A=\frac{2\sqrt{5}}{5}$でもよい．

% 側注2
三平方の定理

\[
a^2+b^2=c^2
\]

% 側注3
$\sin 30^{\circ}=\frac{1}{2}$

% 側注4
$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$

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% 問題I4.1.1の側注：直角三角形の三角比（One More）★
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% 側注1
三平方の定理

\[
a^2+b^2=c^2
\]

% 側注2
$\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

% 側注3
$\cos 60^{\circ}=\frac{1}{2}$

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% 例題I4.1.1の考え方・補足など：直角三角形の三角比（One More）★
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三角比の定義を利用する．$\sin{A}=\frac{a}{c}$

$\cos{A}=\frac{b}{c}$

$\tan{A}=\frac{a}{b}$