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三角形AOPはAO=POの二等辺三角形なので、∠OAP=1/2(180°-Θ)=90°-Θ/2
また
∠OAQ=90°なので
∠PAQ=∠OAQ-∠OAP=90°-(90°-Θ/2)=Θ/2
ありがとうございます♪
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三角形AOPはAO=POの二等辺三角形なので、∠OAP=1/2(180°-Θ)=90°-Θ/2
また
∠OAQ=90°なので
∠PAQ=∠OAQ-∠OAP=90°-(90°-Θ/2)=Θ/2
ありがとうございます♪
Users viewing this question
are also looking at these questions 😉
ありがとうございます♪