Question

7 practice 7 (from unit 4, lesson 6) technology required. this diagram shows several measurements in a triangle. what is the value of ( x )? round your answer to the nearest tenth. type your answer in the box. ______ units how did i do?

Explanation:

Step1: Find the third angle of the triangle

The sum of angles in a triangle is \(180^\circ\). So the angle at \(F\) is \(180^\circ - 40^\circ - 50^\circ = 90^\circ\). Wait, no, wait. Wait, the triangle has angles at \(D\) (40°), at the right angle (50°? Wait, no, the diagram shows a right angle? Wait, no, the angle at the right is 50°? Wait, no, the diagram: point \(D\), then a horizontal side \(x\) to a right angle? Wait, no, the angle at the right is 50°, and angle at \(D\) is 40°, and side \(DF\) is 3. Wait, maybe using the Law of Sines.

Law of Sines: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

In triangle \(D\) (angle 40°), \(F\) (angle let's calculate: 180 - 40 - 50 = 90°? Wait, no, the angle at the right is 50°, and angle at \(D\) is 40°, so angle at \(F\) is \(180 - 40 - 50 = 90^\circ\)? Wait, no, maybe the right angle is a typo? Wait, no, the diagram shows a right angle symbol? Wait, the user's diagram: "50°" with a right angle symbol? Wait, maybe it's a right triangle? Wait, no, angle at \(D\) is 40°, angle at the other end (let's say \(E\)) is 50°, and angle at \(F\) is 90°? Wait, side \(DF\) is 3, opposite angle 50°, and side \(x\) is opposite angle at \(F\) (90°)? Wait, no, let's re-express.

Wait, triangle \(DEF\): angle at \(D\) is 40°, angle at \(E\) is 50°, so angle at \(F\) is \(180 - 40 - 50 = 90^\circ\). So it's a right triangle? Wait, 40 + 50 = 90, so angle at \(F\) is 90°. So side \(DF\) is 3, which is adjacent to angle \(D\) (40°), and side \(x\) is the hypotenuse? Wait, no, Law of Sines: \(\frac{DF}{\sin \angle E} = \frac{DE}{\sin \angle F}\)

So \(DF = 3\), angle at \(E\) is 50°, angle at \(F\) is 90°, so \(\frac{3}{\sin 50^\circ} = \frac{x}{\sin 90^\circ}\)

Since \(\sin 90^\circ = 1\), so \(x = \frac{3}{\sin 50^\circ}\)

Calculate \(\sin 50^\circ \approx 0.7660\)

So \(x = \frac{3}{0.7660} \approx 3.9\)

Wait, let's check again.

Wait, angle at \(D\): 40°, angle at \(E\): 50°, angle at \(F\): 90°. Side \(DF\) is opposite angle \(E\) (50°), side \(DE\) (which is \(x\)) is opposite angle \(F\) (90°). So Law of Sines: \(\frac{DF}{\sin \angle E} = \frac{DE}{\sin \angle F}\)

So \(DF = 3\), \(\angle E = 50^\circ\), \(\angle F = 90^\circ\), so:

\(\frac{3}{\sin 50^\circ} = \frac{x}{\sin 90^\circ}\)

\(\sin 90^\circ = 1\), so \(x = \frac{3}{\sin 50^\circ}\)

\(\sin 50^\circ \approx 0.7660\), so \(x \approx \frac{3}{0.7660} \approx 3.9\)

Wait, but let's confirm. Alternatively, if angle at \(F\) is 90°, then it's a right triangle, so \(\cos(40^\circ) = \frac{DF}{DE}\), so \(DE = \frac{DF}{\cos(40^\circ)}\). Wait, \(DF\) is 3, adjacent to angle \(D\) (40°), so \(DE = x = \frac{3}{\cos(40^\circ)}\). \(\cos(40^\circ) \approx 0.7660\), so \(x \approx 3 / 0.7660 \approx 3.9\). Yes, same result.

So using Law of Sines or right triangle trigonometry.

Step1: Identify angles and sides

Triangle \(DEF\): \(\angle D = 40^\circ\), \(\angle E = 50^\circ\), so \(\angle F = 90^\circ\). Side \(DF = 3\) (opposite \(\angle E = 50^\circ\)), side \(x = DE\) (opposite \(\angle F = 90^\circ\)).

Step2: Apply Law of Sines

\(\frac{DF}{\sin \angle E} = \frac{DE}{\sin \angle F}\)

Substitute values: \(\frac{3}{\sin 50^\circ} = \frac{x}{\sin 90^\circ}\)

Since \(\sin 90^\circ = 1\), \(x = \frac{3}{\sin 50^\circ}\)

Step3: Calculate \(\sin 50^\circ\)

\(\sin 50^\circ \approx 0.7660\)

Step4: Compute \(x\)

\(x = \frac{3}{0.7660} \approx 3.9\) (rounded to nearest tenth)

Answer:

3.9