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 \dots

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

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

l o o k t h i s p i c t u r e s i r

Commented by jagoll last updated on 05/Jan/20

thanks you sir

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

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

A P = \sqrt{4^{2} + 8^{2}} = 4 \sqrt{5}

T P = \sqrt{6^{2} - 4^{2}} = 2 \sqrt{5}

A Q = Q T = \frac{6}{2} = 3

\cos A = \frac{6^{2} + {(4 \sqrt{5})}^{2} - {(2 \sqrt{5})}^{2}}{2 \times 6 \times 4 \sqrt{5}} = \frac{2}{\sqrt{5}}

{P Q}^{2} = 3^{2} + {(4 \sqrt{5})}^{2} - 2 \times 3 \times 4 \sqrt{5} \times \cos A = 41

\cos α = \frac{41 + {(2 \sqrt{5})}^{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

g o o d s i r . i^{'} m t o s o l v e b y v e c t o r . w h a t r e s u l t y o u g o t s i r ?

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

i g o t \cos α = \frac{13}{\sqrt{205}} s i r

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

\cos α = \frac{P \bar{T} . P \bar{Q}}{∣ P \bar{T} ∣ ∣ P \bar{Q} ∣}

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

e r r o r s i r l a s t l i n e = \frac{52}{4 \sqrt{205}} = \frac{13}{\sqrt{205}} \neq \frac{3 \sqrt{205}}{83}

Commented by jagoll last updated on 05/Jan/20

thanks you sir

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

t h a n k s s i r! i h a v e c o r r e c t e d .

Commented by Tawa11 last updated on 29/Dec/21

Great sir

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