Question

select the correct statement in the list.
mrs. henry gave her students the information in the table and asked them to find the unknown parts of δdef, if possible
δdef
d = 7, e = 10, m∠e = 38°
which statement is true?

• there is no set of values for f, m∠d, and m∠f that will create a triangle.
• there is only one set of values for f, m∠d, and m∠f that will create a triangle.
• there are exactly two distinct sets of values for f, m∠d, and m∠f that will create a triangle.
• there are infinitely many sets of values for f, m∠d, and m∠f that will create a triangle.
• there is not enough information to determine whether any sets of values for f, m∠d, and m∠f exist that will create a triangle.

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that for a triangle with sides \(a\), \(b\), \(c\) and opposite angles \(A\), \(B\), \(C\) respectively, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle DEF\), we have side \(d = 7\) (opposite \(\angle D\)), side \(e = 10\) (opposite \(\angle E=38^\circ\)). So by the Law of Sines, \(\frac{d}{\sin D}=\frac{e}{\sin E}\), which gives \(\frac{7}{\sin D}=\frac{10}{\sin 38^\circ}\).

Step2: Solve for \(\sin D\)

From \(\frac{7}{\sin D}=\frac{10}{\sin 38^\circ}\), we can cross - multiply to get \(\sin D=\frac{7\sin 38^\circ}{10}\). Calculate \(\sin 38^\circ\approx0.6157\), then \(\sin D=\frac{7\times0.6157}{10}=\frac{4.3099}{10} = 0.43099\).

Step3: Analyze the value of \(\sin D\)

Since \(\sin D = 0.43099\) and \(0<\sin D<1\), and we know that if \(\sin\theta = k\) where \(0 < k<1\), then \(\theta\) can have two solutions in the interval \((0^\circ,180^\circ)\): one acute angle \(\theta_1=\arcsin(k)\) and one obtuse angle \(\theta_2 = 180^\circ-\arcsin(k)\), provided that \(\theta_2<180^\circ - \angle E\) (to form a valid triangle). Let's check the acute angle first: \(\arcsin(0.43099)\approx25.5^\circ\). The obtuse angle would be \(180 - 25.5=154.5^\circ\). Now, check if the obtuse angle is valid. The sum of angles in a triangle is \(180^\circ\). If \(\angle D = 154.5^\circ\) and \(\angle E = 38^\circ\), then \(\angle D+\angle E=154.5 + 38=192.5>180^\circ\), which is not possible. So there is only one valid acute angle for \(\angle D\).

Step4: Analyze the number of triangles

Once we have \(\angle D\), we can find \(\angle F=180^\circ-\angle D - \angle E\), and then use the Law of Sines again to find side \(f\) (since \(\frac{f}{\sin F}=\frac{e}{\sin E}\)). Since there is only one valid value for \(\angle D\), there will be only one set of values for \(f\), \(m\angle D\), and \(m\angle F\) that will create a triangle. Wait, no, wait. Wait, actually, when we use the Law of Cosines to find side \(f\): \(d^{2}=e^{2}+f^{2}-2ef\cos D\)? No, wait, the given is \(d = 7\), \(e = 10\), \(\angle E=38^\circ\). Let's use the Law of Cosines to find side \(f\) in terms of the other sides and angle. Wait, the Law of Cosines for \(\angle E\) is \(d^{2}=e^{2}+f^{2}-2ef\cos E\). So \(7^{2}=10^{2}+f^{2}-2\times10\times f\times\cos 38^\circ\).

Step5: Rearrange the quadratic equation

\(49 = 100+f^{2}-20f\times0.7880\) (since \(\cos 38^\circ\approx0.7880\)). Rearranging gives \(f^{2}-15.76f + 100 - 49=0\), so \(f^{2}-15.76f + 51 = 0\).

Step6: Calculate the discriminant of the quadratic

For a quadratic equation \(ax^{2}+bx + c = 0\), the discriminant is \(\Delta=b^{2}-4ac\). Here, \(a = 1\), \(b=- 15.76\), \(c = 51\). \(\Delta=(-15.76)^{2}-4\times1\times51=248.3776 - 204 = 44.3776>0\). Wait, but earlier with the Law of Sines, we thought there was one triangle, but the quadratic has two solutions? Wait, no, let's go back. The Law of Sines: \(\sin D=\frac{7\sin38^\circ}{10}\approx\frac{7\times0.6157}{10}\approx0.431\). The value of \(\sin D\approx0.431\), and since \(d = 7

Answer:

There is only one set of values for \(f\), \(m\angle D\), and \(m\angle F\) that will create a triangle.