Question

consider the following figure.
(note that the figure is not drawn to scale.)
(a) order the angle measures ( mangle c ), ( mangle d ), and ( mangle ced ) from least to greatest.
( mangle square < mangle square < mangle square )
(b) order the side lengths ( ef ), ( fg ), and ( eg ) from least to greatest.
( square < square < square )

Explanation:

Step1: Solve part (a) - Angle Order

First, in triangle \( CED \), we know that the sum of angles in a triangle is \( 180^\circ \). Given \( \angle CED = 66^\circ \) (vertical angles with \( \angle FEG \), but wait, no, in triangle \( CED \), sides \( CE = 16 \), \( DE = 6 \), \( CD = 14.63 \)? Wait, no, first, let's find \( m\angle C \), \( m\angle D \), and \( m\angle CED \). Wait, \( \angle CED = 66^\circ \) (given as the angle at E in triangle CED). Then, in triangle \( CED \), using the Law of Sines or just angle - sum. Wait, alternatively, in triangle \( FEG \), \( \angle F = 71^\circ \), and \( \angle FEG \) is vertical to \( \angle CED \), so \( \angle FEG = 66^\circ \), so in triangle \( FEG \), \( \angle G = 180 - 71 - 66 = 43^\circ \). But maybe we can use triangle \( CED \): let's find \( m\angle C \) and \( m\angle D \). Wait, sides: \( CE = 16 \), \( DE = 6 \), \( CD = 14.63 \). Using the Law of Cosines on triangle \( CED \) to find angles? Wait, no, maybe it's easier to use the fact that in a triangle, larger side is opposite larger angle. In triangle \( CED \), side \( CE = 16 \), \( DE = 6 \), \( CD = 14.63 \). So side opposite \( \angle D \) is \( CE = 16 \), side opposite \( \angle C \) is \( DE = 6 \), side opposite \( \angle CED \) is \( CD = 14.63 \). So since \( 6<14.63<16 \), the angles opposite them: \( \angle C \) (opposite \( DE = 6 \)), \( \angle CED \) (opposite \( CD = 14.63 \)), \( \angle D \) (opposite \( CE = 16 \)). So \( m\angle C < m\angle CED < m\angle D \)? Wait, no, wait: side \( DE = 6 \) is opposite \( \angle C \), side \( CE = 16 \) is opposite \( \angle D \), side \( CD = 14.63 \) is opposite \( \angle CED \). So since \( DE = 6 \) is the shortest side, \( \angle C \) (opposite) is the smallest angle. \( CD = 14.63 \) is middle, so \( \angle CED \) is middle. \( CE = 16 \) is longest, so \( \angle D \) is largest. Wait, but also, we have \( \angle CED = 66^\circ \) (given). Let's calculate \( m\angle C \) and \( m\angle D \) using the Law of Sines: \( \frac{\sin C}{DE}=\frac{\sin D}{CE}=\frac{\sin CED}{CD} \). So \( \frac{\sin C}{6}=\frac{\sin D}{16}=\frac{\sin 66^\circ}{14.63} \). \( \sin 66^\circ\approx0.9135 \), so \( \frac{0.9135}{14.63}\approx0.0624 \). Then \( \sin C=\frac{6\times0.0624}{1}\approx0.3744 \), so \( m\angle C\approx22^\circ \) (wait, that can't be, maybe I made a mistake). Wait, no, maybe the triangle is not \( CED \) with sides 16, 6, 14.63. Wait, the figure: \( CE = 16 \), \( DE = 6 \), and \( CD = 14.63 \). Let's check: \( 6 + 14.63>16 \) (6 + 14.63 = 20.63>16), \( 16+6 > 14.63 \) (22>14.63), \( 16 + 14.63>6 \), so it's a valid triangle. Now, using Law of Cosines for \( \angle C \): \( \cos C=\frac{CE^{2}+CD^{2}-DE^{2}}{2\cdot CE\cdot CD}=\frac{16^{2}+14.63^{2}-6^{2}}{2\times16\times14.63}=\frac{256 + 214.0369-36}{468.16}=\frac{434.0369}{468.16}\approx0.927 \), so \( m\angle C\approx21.9^\circ\approx22^\circ \). For \( \angle D \): \( \cos D=\frac{DE^{2}+CD^{2}-CE^{2}}{2\cdot DE\cdot CD}=\frac{6^{2}+14.63^{2}-16^{2}}{2\times6\times14.63}=\frac{36 + 214.0369 - 256}{175.56}=\frac{-4.9631}{175.56}\approx - 0.0282 \), so \( m\angle D\approx91.6^\circ \). And \( m\angle CED = 66^\circ \). So now we have \( m\angle C\approx22^\circ \), \( m\angle CED = 66^\circ \), \( m\angle D\approx91.6^\circ \). So the order from least to greatest is \( m\angle C < m\angle CED < m\angle D \).

Step2: Solve part (b) - Side Order in Triangle \( FEG \)

In triangle \( FEG \), we know \( \angle F = 71^\circ \), \( \angle FEG = 66^\circ \) (vertical to \( \angle CED \)), so \( \angle G=180 - 71 - 66 = 43^\circ \). In a triangle, the larger angle is opposite the larger side. So angles: \( \angle G = 43^\circ \), \( \angle FEG = 66^\circ \), \( \angle F = 71^\circ \). So sides opposite: \( EF \) is opposite \( \angle G \), \( FG \) is opposite \( \angle FEG \), \( EG \) is opposite \( \angle F \). So since \( 43^\circ<66^\circ<71^\circ \), the sides opposite: \( EF \) (opposite \( 43^\circ \)), \( FG \) (opposite \( 66^\circ \)), \( EG \) (opposite \( 71^\circ \)). So the order from least to greatest is \( EF < FG < EG \).

Answer:

(a) \( m\angle C < m\angle CED < m\angle D \)
(b) \( EF < FG < EG \)