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: Wait, \( 36 - 9 = 27 \), but \( 27\times38 = 9\times3\times38 \), but wait, maybe I made a mistake. Wait, \( (6\sqrt{38})^2 = 6^2\times(\sqrt{38})^2 = 36\times38 \), \( (3\sqrt{38})^2 = 9\times38 \). So \( 36\times38 - 9\times38 = (36 - 9)\times38 = 27\times38 \). But \( 27 = 9\times3 \), so \( \sqrt{27\times38}=\sqrt{9\times3\times38}=3\sqrt{114} \)? Wait, no, wait, maybe another way. Wait, notice that \( DE = \frac{1}{2}DF \), since \( 3\sqrt{38}=\frac{1}{2}\times6\sqrt{38} \). So this is a 30-60-90 triangle? Wait, in a 30-60-90 triangle, the sides are in ratio \( 1 : \sqrt{3} : 2 \). Here, \( DE \) is half of \( DF \), so \( \angle F = 30^\circ \), \( \angle D = 60^\circ \), and \( EF = DE\times\sqrt{3} \). Let's check: \( DE = 3\sqrt{38} \), so \( EF = 3\sqrt{38}\times\sqrt{3}=3\sqrt{114} \)? Wait, no, wait, if \( DE \) is adjacent to \( \angle F \), and \( DF \) is hypotenuse, then \( \cos\angle F=\frac{EF}{DF} \), \( \sin\angle F=\frac{DE}{DF} \). \( \sin\angle F=\frac{3\sqrt{38}}{6\sqrt{38}}=\frac{1}{2} \), so \( \angle F = 30^\circ \), because \( \sin30^\circ=\frac{1}{2} \). Then \( \angle D = 90^\circ - 30^\circ = 60^\circ \). Then \( EF \): since \( \cos\angle F=\frac{EF}{DF} \), \( \cos30^\circ=\frac{\sqrt{3}}{2} \), so \( EF = DF\times\cos30^\circ = 6\sqrt{38}\times\frac{\sqrt{3}}{2}=3\sqrt{38}\times\sqrt{3}=3\sqrt{114} \)? Wait, but let's recalculate Pythagorean theorem. \( EF^2 + (3\sqrt{38})^2 = (6\sqrt{38})^2 \). So \( EF^2 = 36\times38 - 9\times38 = 27\times38 = 9\times3\times38 = 9\times114 \), so \( EF = \sqrt{9\times114}=3\sqrt{114} \). Wait, but \( 3\times38 = 114 \), yes. So that's correct.

Step2: Find \( m\angle F \)

Using sine: \( \sin\angle F = \frac{DE}{DF} = \frac{3\sqrt{38}}{6\sqrt{38}} = \frac{1}{2} \). So \( \angle F = 30^\circ \), since \( \sin30^\circ=\frac{1}{2} \).

Step3: Find \( m\angle D \)

In a right triangle, the two acute angles sum to \( 90^\circ \). So \( m\angle D = 90^\circ - m\angle F = 90^\circ - 30^\circ = 60^\circ \).

Answer:

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

Wait, wait, no, wait: Wait, \( 27\times38 = 9\times3\times38 = 9\times114 \), so \( \sqrt{9\times114}=3\sqrt{114} \). Alternatively, maybe I made a mistake in the ratio. Wait, \( DE = 3\sqrt{38} \), \( DF = 6\sqrt{38} \), so \( DE = \frac{1}{2}DF \), so triangle is 30-60-90, with \( \angle F = 30^\circ \), \( \angle D = 60^\circ \), and \( EF = DE\times\sqrt{3} = 3\sqrt{38}\times\sqrt{3}=3\sqrt{114} \). Yes, that's correct.