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Question Number 189807 by normans last updated on 22/Mar/23

	Answered by a.lgnaoui last updated on 22/Mar/23

	$A n g l e e n t r e v e r t i c a l e e t p q (α = a r c \sin \frac{1}{2} = \frac{π}{6})$ $A n g l e e n t r e E q e t q r e s t \frac{π}{4}$ $\frac{π}{6} + X + \frac{π}{4} = π \Rightarrow X = \frac{7 π}{12}$
	Commented by a.lgnaoui last updated on 22/Mar/23

	Answered by cortano12 last updated on 22/Mar/23
	$R(2,1,0) ,Q(2,0,1),P(0,1,2) { ((QR^(→) = (0,1,−1))),((QP^(→) =(−2,1,1))) :} ⇒cos X = ((QR^(→) .QP^(→) )/(∣QR^(→) ∣ .∣QP^(→) ∣)) ⇒ cos X=((0+1−1)/( (√2).(√6)))= 0 ⇒X = 90°$
	$R (2, 1, 0), Q (2, 0, 1), P (0, 1, 2)$ ${\begin{cases} \vec{QR} = (0, 1, - 1) \\ \vec{QP} = (- 2, 1, 1) \end{cases}$ $\Rightarrow \cos X = \frac{\vec{QR} . \vec{QP}}{∣ \vec{QR} ∣ . ∣ \vec{QP} ∣}$ $\Rightarrow \cos X = \frac{0 + 1 - 1}{\sqrt{2} . \sqrt{6}} = 0$ $\Rightarrow X = 90 °$
	Answered by mr W last updated on 22/Mar/23

	$s a y e d g e l e n g t h o f c u b e i s 2 .$ $q r = \sqrt{2}$ $q p = \sqrt{6}$ ${q r}^{2} + {q p}^{2} = 2 + 6 = 8$ $p r = 2 \sqrt{2}$ ${p r}^{2} = 8$ ${p r}^{2} = {q r}^{2} + {q p}^{2}$ $\Rightarrow Δ p q r i s r i g h t a n g l e d . \Rightarrow x = 90 °$
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