Question

practice 8 (from unit 4, lesson 6)
technology required. here is triangle def.
what is the value of ( y )? round your answer to the nearest tenth if necessary.
type your answer in the box.
______ units

Explanation:

Step1: Identify the triangle type and use the Law of Sines

In triangle \( DEF \), we know two angles: \( \angle D = 42^\circ \), \( \angle E = 90^\circ \) (since there's a right angle symbol), so we can find \( \angle F = 180^\circ - 42^\circ - 90^\circ = 48^\circ \)? Wait, no, wait, the angle at \( E \) is adjacent to the side of length 7, and the right angle? Wait, maybe I misread. Wait, the triangle has a right angle? Wait, the diagram shows a right angle at \( E \)? Wait, the side \( DE = 7 \), angle at \( D \) is \( 42^\circ \), angle at \( E \) is \( 90^\circ \)? Wait, no, the angle at \( E \) is labeled with a right angle? Wait, maybe it's a right triangle? Wait, no, the angle at \( E \) is \( 48^\circ \)? Wait, no, the diagram: point \( D \), \( E \), \( F \). \( DE = 7 \), angle at \( D \) is \( 42^\circ \), angle at \( E \) is \( 48^\circ \)? Wait, no, the right angle symbol? Wait, maybe it's a right triangle? Wait, no, the problem says "Technology required. Here is triangle \( DEF \)." Let's re-examine. The angles: \( \angle D = 42^\circ \), \( \angle E = 90^\circ \)? Wait, the right angle symbol is at \( E \)? Wait, the side \( DE = 7 \), and we need to find \( y = DF \)? Wait, no, maybe \( y \) is \( DF \) or \( EF \)? Wait, let's use the Law of Sines. In any triangle, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \). Let's denote: \( \angle D = 42^\circ \), \( \angle E = 90^\circ \) (right angle), so \( \angle F = 180 - 42 - 90 = 48^\circ \). Wait, but the side opposite \( \angle F \) is \( DE = 7 \), the side opposite \( \angle E \) (90°) is \( DF = y \), and the side opposite \( \angle D \) (42°) is \( EF \). Wait, Law of Sines: \( \frac{DE}{\sin F} = \frac{DF}{\sin E} \). So \( DE = 7 \), \( \angle F = 48^\circ \), \( \angle E = 90^\circ \), so \( \frac{7}{\sin 48^\circ} = \frac{y}{\sin 90^\circ} \). Since \( \sin 90^\circ = 1 \), then \( y = \frac{7}{\sin 48^\circ} \). Wait, no, maybe I mixed up the angles. Wait, let's check the angles again. If \( \angle D = 42^\circ \), \( \angle E = 48^\circ \), then \( \angle F = 180 - 42 - 48 = 90^\circ \). Ah! That makes sense. So triangle \( DEF \) has angles: \( \angle D = 42^\circ \), \( \angle E = 48^\circ \), \( \angle F = 90^\circ \). So it's a right triangle with right angle at \( F \). Then \( DE \) is the hypotenuse? Wait, no, \( DE = 7 \), and we need to find \( y \), which is \( DF \) or \( EF \). Wait, let's clarify the sides. Let's label the triangle: \( \angle D = 42^\circ \), \( \angle E = 48^\circ \), \( \angle F = 90^\circ \). So side opposite \( \angle D \) is \( EF \), side opposite \( \angle E \) is \( DF \), and side opposite \( \angle F \) is \( DE = 7 \). So by Law of Sines: \( \frac{DF}{\sin E} = \frac{DE}{\sin F} \). So \( \angle E = 48^\circ \), \( \angle F = 90^\circ \), \( DE = 7 \). So \( \frac{y}{\sin 48^\circ} = \frac{7}{\sin 90^\circ} \). Since \( \sin 90^\circ = 1 \), then \( y = 7 \times \sin 48^\circ \)? Wait, no, that can't be. Wait, maybe I got the angles wrong. Wait, the angle at \( D \) is \( 42^\circ \), angle at \( E \) is \( 90^\circ \), so angle at \( F \) is \( 48^\circ \). Then side \( DE = 7 \) is adjacent to \( \angle D \), and \( y \) is the hypotenuse? Wait, no, in a right triangle, if \( \angle E = 90^\circ \), then \( DE \) and \( EF \) are legs, \( DF \) is hypotenuse. Wait, angle at \( D \) is \( 42^\circ \), so \( \cos(42^\circ) = \frac{DE}{DF} \), so \( DF = \frac{DE}{\cos(42^\circ)} \). \( DE = 7 \), so \( DF = \frac{7}{\cos(42^\circ)} \). Let's calculate that. \( \cos(42^\circ) \approx 0.7431 \), so \( \frac{7}{0.7431} \approx 9.4 \). Wait, but let's check with Law of Sines. Alternatively, if angle at \( F \) is \( 48^\circ \), and side \( DE = 7 \) is opposite \( \angle F \), then \( \frac{y}{\sin 90^\circ} = \frac{7}{\sin 48^\circ} \), so \( y = \frac{7}{\sin 48^\circ} \). \( \sin 48^\circ \approx 0.7431 \), so \( \frac{7}{0.7431} \approx 9.4 \). Yes, that matches. So \( y \approx 9.4 \).

Step2: Calculate the value

Using a calculator, \( \sin 48^\circ \approx 0.7431 \). Then \( y = \frac{7}{\sin 48^\circ} \approx \frac{7}{0.7431} \approx 9.4 \).

Answer:

\( 9.4 \)