Solve question number 21
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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]