MATH SOLVE

4 months ago

Q:
# What is the value of Y?

Accepted Solution

A:

Apply Pythagorean theorem in ∆TMU [tex]\\ \rm\Rrightarrow UT^2=6^2-3^2=36-9=27[/tex] [tex]\\ \rm\Rrightarrow UT=3\sqrt{3}[/tex]Nowin ∆TNU [tex]\\ \rm\Rrightarrow y^2=(3\sqrt)^2+9^2[/tex] [tex]\\ \rm\Rrightarrow y^2=27+81[/tex] [tex]\\ \rm\Rrightarrow y^2=108[/tex] [tex]\\ \rm\Rrightarrow y=6\sqrt{3}[/tex]