Cho hình chóp đều S.ABCD có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpcaaIYaGaamyyaiaacYcacaWGtbGaamyqaiabg2da9iaa % dggadaGcaaqaaiaaiwdaaSqabaaaaa!3F3D! AB = 2a,SA = a\sqrt 5 \) . Góc giữa hai mặt phẳng (SAB) và (ABCD) bằng:
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Lời giải:
Báo saiGọi O là giao điểm của AC và BD
SABCD là hình chóp đều \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4uaiaad+eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaiaadsea % aiaawIcacaGLPaaaaaa!4055! \Rightarrow SO \bot \left( {ABCD} \right)\)
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbGaamyqaiaadkeaaiaawIcacaGLPaaacqGHPiYXdaqadaqaaiaa % dgeacaWGcbGaam4qaiaadseaaiaawIcacaGLPaaacqGH9aqpdaGada % qaaiaadgeacaWGcbaacaGL7bGaayzFaaaaaa!44EB! \left( {SAB} \right) \cap \left( {ABCD} \right) = \left\{ {AB} \right\}\)
Gọi M là trung điểm của AB.
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaad2 % eacqGHLkIxcaWGbbGaamOqamaabmaabaGaam4taiaad2eacaGGVaGa % ai4laiaadgeacaWGebGaaiilaiaadgeacaWGebGaeyyPI4Laamyqai % aadkeaaiaawIcacaGLPaaaaaa!4679! OM \bot AB\left( {OM//AD,AD \bot AB} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad2 % eacqGHLkIxcaWGbbGaamOqaaaa!3ADC! SM \bot AB\) do \(\Delta SAB\) là tam giác cân tại S.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taey % iiIa9aaeWaaeaadaqadaqaaiaadofacaWGbbGaamOqaaGaayjkaiaa % wMcaaiaacYcadaqadaqaaiaadgeacaWGcbGaam4qaiaadseaaiaawI % cacaGLPaaaaiaawIcacaGLPaaacqGH9aqpcqGHGic0daqadaqaaiaa % dofacaWGnbGaaiilaiaad+eacaWGnbaacaGLOaGaayzkaaGaeyypa0 % JaeyiiIaTaam4uaiaad2eacaWGpbaaaa!520C! \Rightarrow \angle \left( {\left( {SAB} \right),\left( {ABCD} \right)} \right) = \angle \left( {SM,OM} \right) = \angle SMO\)
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad2 % eacqGH9aqpdaGcaaqaaiaadofacaWGbbWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaamytaiaadgeadaahaaWcbeqaaiaaikdaaaaabeaaki % abg2da9maakaaabaGaaGynaiaadggadaahaaWcbeqaaiaaikdaaaGc % cqGHsislcaWGHbWaaWbaaSqabeaacaaIYaaaaaqabaGccqGH9aqpca % aIYaGaamyyaaaa!47D9! SM = \sqrt {S{A^2} - M{A^2}} = \sqrt {5{a^2} - {a^2}} = 2a\) (Định lý Pitago)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaad2 % eacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadgeacaWGebGa % eyypa0Jaamyyaaaa!3DA2! OM = \frac{1}{2}AD = a\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3ciGGJbGaai4BaiaacohacaWGtbGaamytaiaad+eacqGH9aqpdaWc % aaqaaiaad+eacaWGnbaabaGaam4uaiaad2eaaaGaeyypa0ZaaSaaae % aacaWGHbaabaGaaGOmaiaadggaaaGaeyypa0ZaaSaaaeaacaaIXaaa % baGaaGOmaaaaaeaacqGHshI3cqGHGic0caWGtbGaamytaiaad+eacq % GH9aqpcaaI2aGaaGimamaaCaaaleqabaGaaGimaaaaaaaa!521A! \begin{array}{l} \Rightarrow \cos SMO = \frac{{OM}}{{SM}} = \frac{a}{{2a}} = \frac{1}{2}\\ \Rightarrow \angle SMO = {60^0} \end{array}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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