shahzeenwaqar1
shahzeenwaqar1 shahzeenwaqar1
  • 15-03-2020
  • Mathematics
contestada

Solve question number 21

Solve question number 21 class=

Respuesta :

iguanamiyagi
iguanamiyagi iguanamiyagi
  • 15-03-2020

Answer:

∝ = 60.50°

Step-by-step explanation:

[tex]AC = \sqrt{AB^{2} + BC^{2} } = \sqrt{128} \\\\AM = \frac{AC}{2} = \frac{\sqrt{128} }{2} \\\\[/tex]

∡VAM = ∝

[tex]tg \alpha = \frac{VM}{AM} = \frac{10}{\frac{\sqrt{128} }{2} } = 1.76776695297 \\\\[/tex]

[tex]\alpha = arctan (1.76776695297) = 60.50[/tex]

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