Question

consider the following figure.
(note that the figure is not drawn to scale.)
(a) order the angle measures ( mangle b ), ( mangle c ), and ( mangle bdc ) from least to greatest.
( manglesquare < manglesquare < manglesquare )
(b) order the side lengths ( de ), ( ef ), and ( df ) from least to greatest.
( square < square < square )

Explanation:

Part (a)

Step1: Find \( m\angle BDC \)

In triangle \( BDC \), we know that the sum of angles in a triangle is \( 180^\circ \). We can find \( m\angle BDC = 68^\circ \) (given as the vertical angle? Wait, no, the angle at \( D \) in triangle \( BDC \) is \( 68^\circ \)? Wait, actually, in triangle \( BDC \), sides \( BD = 10 \), \( DC = 16 \), \( BC = 15.37 \). Wait, maybe we can use the Law of Sines or Law of Cosines, but first, let's recall that in a triangle, larger side is opposite larger angle.

First, in triangle \( BDC \):

• Side \( BD = 10 \), \( DC = 16 \), \( BC = 15.37 \)
• So, order of sides: \( BD < BC < DC \) (since \( 10 < 15.37 < 16 \))
• Therefore, the angles opposite them: \( \angle C \) (opposite \( BD \)), \( \angle B \) (opposite \( DC \)), \( \angle BDC \) (opposite \( BC \))
• Wait, no: in triangle \( BDC \), angle opposite \( BD \) is \( \angle C \), angle opposite \( DC \) is \( \angle B \), angle opposite \( BC \) is \( \angle BDC \)
• So, since \( BD = 10 \) (shortest side) \( \implies \angle C \) (opposite) is smallest
• \( BC = 15.37 \) (middle side) \( \implies \angle BDC \) (opposite) is middle
• \( DC = 16 \) (longest side) \( \implies \angle B \) (opposite) is largest? Wait, no, wait: \( BD = 10 \), \( BC = 15.37 \), \( DC = 16 \). So \( BD < BC < DC \). Therefore, angles opposite: \( \angle C \) (opposite \( BD \)) < \( \angle BDC \) (opposite \( BC \)) < \( \angle B \) (opposite \( DC \))? Wait, no, wait: angle opposite \( BD \) is \( \angle C \), angle opposite \( DC \) is \( \angle B \), angle opposite \( BC \) is \( \angle BDC \). So:

If \( BD < BC < DC \), then \( \angle C < \angle BDC < \angle B \)? Wait, no, that can't be, because \( \angle BDC \) is given as \( 68^\circ \)? Wait, the angle at \( D \) in triangle \( BDC \) is \( 68^\circ \), so \( m\angle BDC = 68^\circ \). Then, sum of angles in triangle \( BDC \): \( \angle B + \angle C + \angle BDC = 180^\circ \implies \angle B + \angle C = 180 - 68 = 112^\circ \)

Now, also, in triangle \( EDF \) and \( BDC \), maybe vertical angles? Wait, \( \angle EDF = \angle BDC = 68^\circ \) (vertical angles). In triangle \( EDF \), angle at \( E \) is \( 74^\circ \), so angle at \( F \) is \( 180 - 74 - 68 = 38^\circ \). But maybe that's not needed here.

Wait, back to triangle \( BDC \): sides \( BD = 10 \), \( DC = 16 \), \( BC = 15.37 \). So, using Law of Cosines to find angles:

For \( \angle B \): \( \cos B = \frac{BD^2 + BC^2 - DC^2}{2 \cdot BD \cdot BC} = \frac{10^2 + 15.37^2 - 16^2}{2 \cdot 10 \cdot 15.37} \)

Calculate numerator: \( 100 + 236.2369 - 256 = 100 + 236.2369 = 336.2369 - 256 = 80.2369 \)

Denominator: \( 2 \cdot 10 \cdot 15.37 = 307.4 \)

So \( \cos B = \frac{80.2369}{307.4} \approx 0.261 \implies \angle B \approx \arccos(0.261) \approx 74.8^\circ \)

For \( \angle C \): \( \cos C = \frac{DC^2 + BC^2 - BD^2}{2 \cdot DC \cdot BC} = \frac{16^2 + 15.37^2 - 10^2}{2 \cdot 16 \cdot 15.37} \)

Numerator: \( 256 + 236.2369 - 100 = 256 + 236.2369 = 492.2369 - 100 = 392.2369 \)

Denominator: \( 2 \cdot 16 \cdot 15.37 = 491.84 \)

So \( \cos C = \frac{392.2369}{491.84} \approx 0.797 \implies \angle C \approx \arccos(0.797) \approx 37.2^\circ \)

And \( \angle BDC = 68^\circ \) (given as the angle at \( D \))

So now, order of angles: \( \angle C \approx 37.2^\circ \), \( \angle BDC = 68^\circ \), \( \angle B \approx 74.8^\circ \)

Therefore, \( m\angle C < m\angle BDC < m\angle B \)

Step2: Confirm with side-angle relationship

In triangle \( BDC \), side \( BD = 10 \) (opposite \( \angle C \)), side \( DC = 16 \) (opposite \( \angle B \)), sid…

Step1: Analyze triangle \( EDF \)

In triangle \( EDF \), we know angle at \( E \) is \( 74^\circ \), angle at \( D \) is \( \angle EDF = \angle BDC = 68^\circ \) (vertical angles), so angle at \( F \) is \( 180 - 74 - 68 = 38^\circ \)

So angles in triangle \( EDF \): \( \angle F = 38^\circ \), \( \angle D = 68^\circ \), \( \angle E = 74^\circ \)

Step2: Order the angles

So order of angles: \( \angle F < \angle D < \angle E \)

Step3: Relate angles to sides (Law of Sines)

In triangle \( EDF \), the sides opposite these angles:

• Side opposite \( \angle F \) is \( DE \)
• Side opposite \( \angle D \) is \( EF \)
• Side opposite \( \angle E \) is \( DF \)

Since in a triangle, larger angle is opposite larger side, so:

• \( \angle F \) (smallest angle) \( \implies DE \) (opposite) is smallest side
• \( \angle D \) (middle angle) \( \implies EF \) (opposite) is middle side
• \( \angle E \) (largest angle) \( \implies DF \) (opposite) is largest side

Wait, wait: angle \( \angle F = 38^\circ \) (opposite \( DE \)), angle \( \angle D = 68^\circ \) (opposite \( EF \)), angle \( \angle E = 74^\circ \) (opposite \( DF \))

So order of angles: \( \angle F < \angle D < \angle E \)

Therefore, order of sides (opposite angles): \( DE < EF < DF \) (since smaller angle opposite smaller side)

Wait, let's confirm:

• Angle \( \angle F = 38^\circ \) (smallest) \( \implies DE \) (opposite) is smallest
• Angle \( \angle D = 68^\circ \) (middle) \( \implies EF \) (opposite) is middle
• Angle \( \angle E = 74^\circ \) (largest) \( \implies DF \) (opposite) is largest

Yes, that's correct. So sides: \( DE < EF < DF \)

Answer:

s:

(a) \( m\angle C < m\angle BDC < m\angle B \)

(b) \( DE < EF < DF \)