+501 In right triangle $DEF$, we have $\angle D = 25^\circ$, $\angle E = 90^\circ$, and $EF = 9$. Find $DE$ to the nearest tenth. You may use a calculator for this problem.
+526 So here's the problem figure
From figure,
\(tan 25 = {EF \over DE}\)
\(0.47={9 \over DE}\)
\(DE=19.15\)
∴ Length of DE is 19.15 cm.
+526 So here's the problem figure
From figure,
\(tan 25 = {EF \over DE}\)
\(0.47={9 \over DE}\)
\(DE=19.15\)
∴ Length of DE is 19.15 cm.
