𝙖) 𝙏𝙖 𝙘ó: 𝙂ó𝙘 𝘼𝘽𝙃 + 𝙜ó𝙘 𝘽𝘼𝙃 = 90° 𝙈à 𝙜ó𝙘 𝘾𝘼𝙃 + 𝙜ó𝙘 𝘽𝘼𝙃 = 90° => 𝙂ó𝙘 𝘼𝘽𝙃 = 𝙂ó𝙘 𝘾𝘼𝙃 𝙓é𝙩 Δ𝘼𝘽𝙃 𝙫à Δ𝘾𝘼𝙆 𝙘ó: - 𝙂ó𝙘 𝙃 = 𝙂ó𝙘 𝘾 (= 90°) - 𝘼𝘽 = 𝘼𝘾 (𝙏𝙖𝙢 𝙜𝙞á𝙘 𝘼𝘽𝘾 𝙫𝙪ô𝙣𝙜 𝙘â𝙣) - 𝙂ó𝙘 𝘼𝘽𝙃 = 𝙂ó𝙘 𝘾𝘼𝙃 (𝙘𝙢𝙩) => Δ𝘼𝘽𝙃 = Δ𝘾𝘼𝙆 (𝙘ạ𝙣𝙝 𝙝𝙪𝙮ề𝙣-𝙜ó𝙘 𝙣𝙝ọ𝙣) => 𝘽𝙃 = 𝘼𝙆 𝙗) 𝙏𝙖 𝙘ó 𝘽𝙃//𝘾𝙆 (𝘾ù𝙣𝙜 𝙫𝙪ô𝙣𝙜 𝙜ó𝙘 𝙫ớ𝙞 𝘼𝙆) =>𝙂ó𝙘 𝙃𝘽𝙈 = 𝙂ó𝙘 𝙈𝘾𝙆 (𝙎𝙤 𝙡𝙚 𝙩𝙧𝙤𝙣𝙜) (1) 𝙈à 𝙜ó𝙘 𝙈𝘼𝙀 + 𝙜ó𝙘 𝘼𝙀𝙈 = 90° (2) 𝙂ó𝙘 𝙈𝘾𝙆 + 𝙜ó𝙘 𝘾𝙀𝙆 = 90° (3) 𝙑à 𝙜ó𝙘 𝘼𝙀𝙈 = 𝙜ó𝙘 𝘾𝙀𝙆 (2 𝙜ó𝙘 đố𝙞 đỉ𝙣𝙝) (4) 𝙏ừ (2),(3),(4) => 𝙂ó𝙘 𝙈𝘼𝙀 = 𝙂ó𝙘 𝙀𝘾𝙆 (5) 𝙏ừ (1),(5) => 𝙂ó𝙘 𝙃𝘽𝙈 = 𝙂ó𝙘 𝙈𝘼𝙀 𝙏𝙖 𝙘ó 𝘼𝙈 𝙡à 𝙩𝙧𝙪𝙣𝙜 𝙩𝙪𝙮ế𝙣 𝙘ủ𝙖 𝙩𝙖𝙢 𝙜𝙞á𝙘 𝙫𝙪ô𝙣𝙜 𝘼𝘽𝘾 𝙣ê𝙣 𝘼𝙈 = 𝘽𝙈 = 𝙈𝘾 = 1/2 𝘽𝘾 𝙓é𝙩 Δ𝙈𝘽𝙃 𝙫à Δ𝙈𝘼𝙆 𝙘ó: - 𝙈𝘽 = 𝘼𝙈 (𝙘𝙢𝙩) - 𝙂ó𝙘 𝙃𝘽𝙈 = 𝙂ó𝙘 𝙈𝘼𝙆(𝙘𝙢𝙩) - 𝘽𝙃 = 𝘼𝙆 (𝙘𝙢 ở 𝙘â𝙪 𝙖) => Δ𝙈𝘽𝙃 = Δ𝙈𝘼𝙆 (𝙘.𝙜.𝙘) 𝙘) 𝙏𝙝𝙚𝙤 𝙘â𝙪 𝙖, 𝙗 𝙩𝙖 𝙘ó: - 𝘼𝙃 = 𝘾𝙆 - 𝙈𝙃 = 𝙈𝙆 - 𝘼𝙈 = 𝙈𝘾 ⇒ Δ𝘼𝙈𝙃 = Δ𝘾𝙈𝙆 (𝙘.𝙘.𝙘) ⇒ 𝙂ó𝙘 𝘼𝙈𝙃 = 𝙂ó𝙘 𝘾𝙈𝙆 𝙈à 𝙜ó𝙘 𝘼𝙈𝙃 + 𝙜ó𝙘 𝙃𝙈𝘾 = 90° ⇒ 𝙂ó𝙘 𝘾𝙈𝙆 + 𝙜ó𝙘 𝙃𝙈𝘾 = 90° 𝙝𝙖𝙮 𝙜ó𝙘 𝙃𝙈𝙆 = 90° Δ𝙃𝙈𝙆 𝙘ó 𝙈𝙆 = 𝙈𝙃 𝙫à 𝙜ó𝙘 𝙃𝙈𝙆 = 90° ⇒ Δ𝙃𝙈𝙆 𝙫𝙪ô𝙣𝙜 𝙘â𝙣 𝙩ạ𝙞 𝙈 (đ𝙥𝙘𝙢) •—Xin câu trả lời hay nhất 😚✨ #hoidap247 #nuocman208-