問題の解答
検索用コード（LaTeX）
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% 例題A3.2.3：接弦定理（One More）★
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次の図において，Oは円の中心，ATは点Aにおける接線とするとき，角$x$を求めよ．
(1)
(2)
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% 例題A3.2.3の解答（One More）★
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(1) 接弦定理より，$\angle\mathrm{BAT}=\angle\mathrm{ACB}=70^\circ$
よって，$\triangle\mathrm{ABC}$の内角の和は$180^\circ$であるから，
\[
x=180^\circ-(30^\circ+70^\circ)=80^\circ
\]
(2) BTと円Oの交点をCとする．
BCは直径であるから，$\angle\mathrm{CAB}=90^{\circ}$
接弦定理より，$\angle\mathrm{TAC}=\angle\mathrm{ABC}=x$
$\triangle\mathrm{ABT}$の内角の和は$180^\circ$であるから，
\[
x+(x+90^{\circ})+40^{\circ}=180^{\circ}
\]
よって，$x=25^{\circ}$
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% 問題A3.2.3：接弦定理（One More）★
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右の図において，ATは点Aにおける接線とするとき，角$x$を求めよ．
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% 問題A3.2.3の解答（One More）★
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四角形ABCDは円に内接するから，
\[
\angle\mathrm{BAD}+\angle\mathrm{BCD}=180^{\circ}
\]
したがって，$\angle\mathrm{BAD}=180^{\circ}-110^{\circ}=70^{\circ}$
接弦定理より，$\angle\mathrm{BAT}=\angle\mathrm{ADB}=65^\circ$
よって，$\triangle\mathrm{ABD}$の内角の和は$180^\circ$であるから，
\[
x=180^\circ-(70^\circ+65^\circ)=45^\circ
\]
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