Question

solve the right triangle.

write your answers in simplified, rationalized form. do not round.
ef =
m∠f =
m∠d =

Explanation:

Step1: Find EF using Pythagorean theorem

In right triangle \( DEF \), \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), and \( \angle E = 90^\circ \). By Pythagorean theorem, \( EF^2 + DE^2 = DF^2 \). So \( EF^2 = DF^2 - DE^2 \). Substitute values: \( EF^2=(6\sqrt{38})^2-(3\sqrt{38})^2 = 36\times38 - 9\times38 = (36 - 9)\times38 = 27\times38 \). Then \( EF=\sqrt{27\times38}=\sqrt{9\times3\times38}=3\sqrt{114} \)? Wait, no, wait: \( 6\sqrt{38} \) squared is \( 36\times38 \), \( 3\sqrt{38} \) squared is \( 9\times38 \), so difference is \( 27\times38 \). Wait, but maybe using trigonometry or noticing the ratio. Wait, \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), so \( \cos D=\frac{DE}{DF}=\frac{3\sqrt{38}}{6\sqrt{38}}=\frac{1}{2} \), so \( \angle D = 60^\circ \), then \( \angle F = 30^\circ \), and \( EF = DE \times \tan D = 3\sqrt{38} \times \sqrt{3}=3\sqrt{114} \)? Wait, no, wait: Pythagorean theorem: \( EF = \sqrt{DF^2 - DE^2} = \sqrt{(6\sqrt{38})^2 - (3\sqrt{38})^2} = \sqrt{36\times38 - 9\times38} = \sqrt{27\times38} = \sqrt{9\times3\times38} = 3\sqrt{114} \)? Wait, no, 27 is 9*3, so \( \sqrt{9\times3\times38}=3\sqrt{114} \). Alternatively, since \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), so the ratio of \( DE \) to \( DF \) is 1/2, so triangle is 30-60-90 triangle. So \( \angle D = 60^\circ \), \( \angle F = 30^\circ \), and \( EF = DE \times \sqrt{3} = 3\sqrt{38} \times \sqrt{3}=3\sqrt{114} \)? Wait, no, in 30-60-90 triangle, the sides are in ratio \( 1 : \sqrt{3} : 2 \). Here, \( DE \) is adjacent to \( \angle D \), \( DF \) is hypotenuse, so \( \cos D = DE/DF = 1/2 \), so \( \angle D = 60^\circ \), then opposite side to \( \angle D \) is \( EF \), so \( EF = DE \times \tan 60^\circ = 3\sqrt{38} \times \sqrt{3}=3\sqrt{114} \). Wait, but let's recalculate Pythagorean:

\( DF^2 = (6\sqrt{38})^2 = 36 \times 38 = 1368 \)

\( DE^2 = (3\sqrt{38})^2 = 9 \times 38 = 342 \)

\( EF^2 = 1368 - 342 = 1026 \)

\( \sqrt{1026} = \sqrt{9 \times 114} = 3\sqrt{114} \). Yes, because 1026 ÷ 9 = 114.

Now, angles: \( \cos \angle D = DE / DF = 3\sqrt{38}/6\sqrt{38} = 1/2 \), so \( \angle D = 60^\circ \), then \( \angle F = 90^\circ - 60^\circ = 30^\circ \).

Step2: Confirm angles

Since \( \cos \angle D = 1/2 \), \( \angle D = 60^\circ \), so \( \angle F = 30^\circ \).

Step3: Summarize

• \( EF = 3\sqrt{114} \) (from Pythagorean theorem: \( \sqrt{DF^2 - DE^2} = \sqrt{1368 - 342} = \sqrt{1026} = 3\sqrt{114} \))
• \( m\angle F = 30^\circ \) (since \( \angle D = 60^\circ \), right triangle, so \( 90 - 60 = 30 \))
• \( m\angle D = 60^\circ \) (from \( \cos \angle D = 1/2 \))

Wait, but let's check again. Wait, \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), so \( DE = \frac{1}{2} DF \), so in right triangle, if one leg is half the hypotenuse, then the angle opposite that leg is 30 degrees. Wait, no: the leg opposite 30 degrees is half the hypotenuse. Here, \( DE \) is adjacent to \( \angle D \), so \( \angle D \) has adjacent side \( DE = 3\sqrt{38} \), hypotenuse \( DF = 6\sqrt{38} \), so \( \cos \angle D = 1/2 \), so \( \angle D = 60^\circ \), so the leg opposite \( \angle D \) is \( EF \), which should be \( DE \times \tan 60^\circ = 3\sqrt{38} \times \sqrt{3} = 3\sqrt{114} \), which matches the Pythagorean result.

Answer:

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

(But let's check the Pythagorean again: \( (6\sqrt{38})^2 = 3638 = 1368 \), \( (3\sqrt{38})^2 = 938 = 342 \), 1368 - 342 = 1026. \( \sqrt{1026} = \sqrt{9*114} = 3\sqrt{114} \). Correct. Angles: \( \cos \theta = 3\sqrt{38}/6\sqrt{38} = 1/2 \), so \( \theta = 60^\circ \), so other angle is 30^\circ. Correct.)