Question-189807

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)

{\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 °

Facebook Tweet Pin