Question

solve the right triangle.

write your answers in simplified, rationalized form. do not round.
ef =

m∠f =

m∠d =

Explanation:

Step1: Use Pythagorean theorem

In right triangle \( DEF \), \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), and \( \angle E = 90^\circ \). By Pythagorean theorem, \( EF=\sqrt{DF^{2}-DE^{2}} \).
Substitute values: \( DF^{2}=(6\sqrt{38})^{2}=36\times38 \), \( DE^{2}=(3\sqrt{38})^{2}=9\times38 \).
So \( EF=\sqrt{36\times38 - 9\times38}=\sqrt{(36 - 9)\times38}=\sqrt{27\times38}=\sqrt{9\times3\times38}=3\sqrt{114} \)? Wait, no, wait: \( 36 - 9 = 27 \)? Wait, no, \( 6^2=36 \), \( 3^2 = 9 \), \( 36-9 = 27 \)? Wait, no, wait, \( DF \) is hypotenuse? Wait, no, in right triangle, hypotenuse is opposite right angle. So \( DF \) is hypotenuse, \( DE \) and \( EF \) are legs. So Pythagorean theorem: \( DE^{2}+EF^{2}=DF^{2} \), so \( EF^{2}=DF^{2}-DE^{2} \).
\( DF^{2}=(6\sqrt{38})^{2}=6^{2}\times(\sqrt{38})^{2}=36\times38 \)
\( DE^{2}=(3\sqrt{38})^{2}=9\times38 \)
So \( EF^{2}=36\times38 - 9\times38=(36 - 9)\times38=27\times38 \)
Wait, but \( 27 = 9\times3 \), so \( EF=\sqrt{9\times3\times38}=3\sqrt{114} \)? Wait, no, wait, maybe I made a mistake. Wait, \( 36 - 9 = 27 \), yes. But wait, alternatively, notice that \( DE=\frac{1}{2}DF \), because \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), so \( DE=\frac{1}{2}DF \). So in a right triangle, if one leg is half the hypotenuse, then the angle opposite that leg is \( 30^\circ \). Wait, \( DE \) is opposite \( \angle F \), so \( \sin(\angle F)=\frac{DE}{DF}=\frac{3\sqrt{38}}{6\sqrt{38}}=\frac{3}{6}=\frac{1}{2} \), so \( \angle F = 30^\circ \), then \( \angle D = 60^\circ \), and \( EF=\sqrt{DF^{2}-DE^{2}} \). Wait, let's recalculate \( EF \):

\( EF=\sqrt{(6\sqrt{38})^{2}-(3\sqrt{38})^{2}}=\sqrt{36\times38 - 9\times38}=\sqrt{(36 - 9)\times38}=\sqrt{27\times38}=\sqrt{9\times3\times38}=3\sqrt{114} \)? Wait, no, \( 38 = 2\times19 \), so can't simplify more? Wait, no, wait, maybe I messed up the legs. Wait, the right angle is at \( E \), so \( DE \) and \( EF \) are legs, \( DF \) is hypotenuse. So \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \). So \( \cos(\angle D)=\frac{DE}{DF}=\frac{3\sqrt{38}}{6\sqrt{38}}=\frac{1}{2} \), so \( \angle D = 60^\circ \), then \( \angle F = 30^\circ \), and \( EF = DE\times\tan(60^\circ)=3\sqrt{38}\times\sqrt{3}=3\sqrt{114} \), or \( EF = DF\times\sin(60^\circ)=6\sqrt{38}\times\frac{\sqrt{3}}{2}=3\sqrt{38}\times\sqrt{3}=3\sqrt{114} \). Wait, but let's check Pythagorean theorem again: \( (3\sqrt{38})^{2}+(3\sqrt{114})^{2}=9\times38 + 9\times114=9\times(38 + 114)=9\times152=1368 \). And \( (6\sqrt{38})^{2}=36\times38 = 1368 \). Yes, that works.

Step2: Find \( m\angle F \)

Since \( \cos(\angle F)=\frac{EF}{DF} \), but we know \( DE=\frac{1}{2}DF \), so in right triangle, if a leg is half the hypotenuse, the angle opposite that leg is \( 30^\circ \). The leg \( DE \) is opposite \( \angle F \), and \( DE=\frac{1}{2}DF \), so \( \angle F = 30^\circ \).

Step3: Find \( m\angle D \)

In right triangle, angles sum to \( 180^\circ \), so \( \angle D = 90^\circ - \angle F = 90^\circ - 30^\circ = 60^\circ \).

Wait, but let's re - check \( EF \). Wait, \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \). So \( \sin(\angle D)=\frac{EF}{DF} \), \( \cos(\angle D)=\frac{DE}{DF}=\frac{3\sqrt{38}}{6\sqrt{38}}=\frac{1}{2} \), so \( \angle D = 60^\circ \), then \( \sin(60^\circ)=\frac{EF}{DF} \), so \( EF = DF\times\sin(60^\circ)=6\sqrt{38}\times\frac{\sqrt{3}}{2}=3\sqrt{38}\times\sqrt{3}=3\sqrt{114} \). Yes, that's correct.

Answer:

\( EF = 3\sqrt{114} \)
\( m\angle F = 30 \)
\( m\angle D = 60 \)