given-T-ABCD-is-pyramid-with-AB-BC-8-and-AT-6-P-is-midpoint-BC-Q-is-midpoint-AT-If-is-the-angle-between-TP-and-PQ-then-cos-is-

Question Number 77272 by jagoll last updated on 05/Jan/20

given T.ABCD is pyramid with AB = BC = 8 and AT = 6 . P is midpoint BC, Q is midpoint AT. If α is the angle between TP and PQ then cos α is ...

$$\mathrm{given}\:\mathrm{T}.\mathrm{ABCD}\:\mathrm{is}\:\mathrm{pyramid}\: \\ $$$$\mathrm{with}\:\mathrm{AB}\:=\:\mathrm{BC}\:=\:\mathrm{8}\:\mathrm{and}\:\mathrm{AT}\:=\:\mathrm{6} \\ $$$$\:.\:\mathrm{P}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{BC},\:\mathrm{Q}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{AT}. \\ $$$$\mathrm{If}\:\alpha\:\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{TP}\:\mathrm{and} \\ $$$$\mathrm{PQ}\:\mathrm{then}\:\mathrm{cos}\:\alpha\:\mathrm{is}\:… \\ $$

Commented by john santu last updated on 05/Jan/20

Commented by john santu last updated on 05/Jan/20

$${look}\:{this}\:{picture}\:{sir} \\ $$

Commented by jagoll last updated on 05/Jan/20

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 05/Jan/20

Commented by mr W last updated on 05/Jan/20

AP=(√(4^2 +8^2 ))=4(√5) TP=(√(6^2 −4^2 ))=2(√5) AQ=QT=(6/2)=3 cos A=((6^2 +(4(√5))^2 −(2(√5))^2 )/(2×6×4(√5)))=(2/( (√5))) PQ^2 =3^2 +(4(√5))^2 −2×3×4(√5)×cos A=41 cos α=((41+(2(√5))^2 −3^2 )/(2×(√(41))×2(√5)))=((13)/( (√(205))))=((13(√(205)))/(205))

$${AP}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{5}} \\ $$$${TP}=\sqrt{\mathrm{6}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${AQ}={QT}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3} \\ $$$$\mathrm{cos}\:{A}=\frac{\mathrm{6}^{\mathrm{2}} +\left(\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{6}×\mathrm{4}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${PQ}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +\left(\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\mathrm{4}\sqrt{\mathrm{5}}×\mathrm{cos}\:{A}=\mathrm{41} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{41}+\left(\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }{\mathrm{2}×\sqrt{\mathrm{41}}×\mathrm{2}\sqrt{\mathrm{5}}}=\frac{\mathrm{13}}{\:\sqrt{\mathrm{205}}}=\frac{\mathrm{13}\sqrt{\mathrm{205}}}{\mathrm{205}} \\ $$

Commented by john santu last updated on 05/Jan/20

$${good}\:{sir}.\:{i}'{m}\:{to}\:{solve}\:{by}\:{vector}\:.\:{what}\:{result}\:{you}\:{got}\:{sir}? \\ $$

Commented by john santu last updated on 05/Jan/20

$${i}\:{got}\:\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{13}}{\:\sqrt{\mathrm{205}\:}}\:{sir} \\ $$

Commented by john santu last updated on 05/Jan/20

cos α = ((PT^� . PQ^� )/(∣PT^� ∣ ∣PQ^� ∣))

$$\mathrm{cos}\:\alpha\:=\:\frac{{P}\bar {{T}}\:.\:{P}\bar {{Q}}}{\mid{P}\bar {{T}}\mid\:\mid{P}\bar {{Q}}\mid} \\ $$

Commented by john santu last updated on 05/Jan/20

error sir last line = ((52)/(4(√(205)))) = ((13)/( (√(205)))) ≠ ((3(√(205)))/(83))

$${error}\:{sir}\:{last}\:{line}\:=\:\frac{\mathrm{52}}{\mathrm{4}\sqrt{\mathrm{205}}}\:=\:\frac{\mathrm{13}}{\:\sqrt{\mathrm{205}}}\:\neq\:\frac{\mathrm{3}\sqrt{\mathrm{205}}}{\mathrm{83}} \\ $$$$ \\ $$

Commented by jagoll last updated on 05/Jan/20