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

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

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

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

$$lookthispicturesir$$

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

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

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

$$goodsir.i^{\prime }mtosolvebyvector.whatresultyougotsir?$$

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

$$igot\cos \alpha =\frac{13}{\sqrt{205}}sir$$

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

$$\cos \alpha =\frac{P\overline{T}.P\overline{Q}}{\mid P\overline{T}\mid \mid P\overline{Q}\mid }$$

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

$$errorsirlastline=\frac{52}{4\sqrt{205}}=\frac{13}{\sqrt{205}}\ne \frac{3\sqrt{205}}{83}$$$$$$

Commented by jagoll last updated on 05/Jan/20

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

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

$$thankssir!ihavecorrected.$$

Commented by Tawa11 last updated on 29/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

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