LeithQ334047 LeithQ334047
  • 03-11-2022
  • Mathematics
contestada

it wants me to solve for the other leg and for the hypotenuse of the 45-45-90 triangle

it wants me to solve for the other leg and for the hypotenuse of the 454590 triangle class=

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WrennaZ253040 WrennaZ253040
  • 03-11-2022

Given:

In the given 45-45-90 triangle,

Use the tan ratio,

[tex]\begin{gathered} \tan 45^{\circ}=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \tan 45^{\circ}=\frac{v}{7} \\ 1=\frac{v}{7} \\ v=7 \end{gathered}[/tex]

Use the cosine ratio,

[tex]\begin{gathered} \cos 45^{\circ}=\frac{adjacent\text{ side}}{\text{hypotenuse}} \\ \cos 45^{\circ}=\frac{7}{u} \\ \frac{1}{\sqrt[]{2}}=\frac{7}{u} \\ u=\frac{7}{\sqrt[]{2}} \\ u=7\sqrt[]{2} \end{gathered}[/tex]

Answer: option a)

[tex]u=7\sqrt[]{2},v=7[/tex]

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