Question

question
in \\(\triangle klm\\), \\(l = 9\\) inches, \\(k = 8.6\\) inches and \\(\angle k = 70^\circ\\). find all possible values of \\(\angle l\\), to the nearest tenth of a degree.
answer
attempt 3 out of 3
\\(\odot\\) additional angle \\(\odot\\) no possible angles
\\(^\circ\\)
submit answer
\\(\odot\\) watch video
show examples

Explanation:

Step1: Apply Law of Sines

In \(\triangle KLM\), by the Law of Sines, we have \(\frac{\sin L}{l}=\frac{\sin K}{k}\). We know \(l = 9\) inches, \(k = 8.6\) inches, and \(\angle K=70^{\circ}\). Substitute these values into the formula:
\(\frac{\sin L}{9}=\frac{\sin 70^{\circ}}{8.6}\)

Step2: Solve for \(\sin L\)

First, calculate \(\sin 70^{\circ}\approx0.9397\). Then:
\(\sin L=\frac{9\times\sin 70^{\circ}}{8.6}=\frac{9\times0.9397}{8.6}=\frac{8.4573}{8.6}\approx0.9834\)

Step3: Find possible angles for \(L\)

We know that if \(\sin\theta = x\), then \(\theta=\arcsin(x)\) or \(\theta = 180^{\circ}-\arcsin(x)\) (in the range \(0^{\circ}<\theta<180^{\circ}\) for a triangle angle).
First, calculate \(\arcsin(0.9834)\approx79.6^{\circ}\). Then the other possible angle is \(180^{\circ}-79.6^{\circ}=100.4^{\circ}\). But we need to check if these angles are valid in the triangle.
For \(\angle L = 79.6^{\circ}\), then \(\angle M=180^{\circ}-70^{\circ}-79.6^{\circ}=30.4^{\circ}\) (valid as all angles are positive and sum to \(180^{\circ}\)).
For \(\angle L = 100.4^{\circ}\), then \(\angle M=180^{\circ}-70^{\circ}-100.4^{\circ}=9.6^{\circ}\) (also valid as all angles are positive and sum to \(180^{\circ}\)). But we need to check the side lengths. Since \(l = 9\) and \(k = 8.6\), \(l>k\), so \(\angle L>\angle K\) (since larger side is opposite larger angle). \(\angle K = 70^{\circ}\), so \(\angle L\) should be greater than \(70^{\circ}\). Both \(79.6^{\circ}\) and \(100.4^{\circ}\) are greater than \(70^{\circ}\), but we also need to check if \(\sin L\approx0.9834\) gives two possible angles. Wait, actually, when we use Law of Sines, the ambiguous case (SSA) occurs when \(k < l\) and \(\angle K\) is acute. Here, \(k = 8.6\), \(l = 9\), so \(k<l\), and \(\angle K = 70^{\circ}\) (acute). So there are two possible triangles? Wait, no, when \(k < l\) and \(\angle K\) is acute, then \(\angle L\) is acute or obtuse? Wait, \(\sin L\approx0.9834\), so \(\angle L\approx79.6^{\circ}\) (acute) or \(180 - 79.6=100.4^{\circ}\) (obtuse). But we need to check if \(l\sin K<k\)? Wait, \(l\sin K=9\times\sin70^{\circ}\approx9\times0.9397 = 8.4573\), and \(k = 8.6\). Since \(l\sin K=8.4573<k = 8.6<l = 9\), so there are two possible triangles. But wait, the problem says "Find all possible values of \(\angle L\)". Wait, but let's check the calculation again. Wait, \(\frac{\sin L}{9}=\frac{\sin70^{\circ}}{8.6}\), so \(\sin L=\frac{9\sin70^{\circ}}{8.6}\approx\frac{9\times0.9396926}{8.6}\approx\frac{8.4572334}{8.6}\approx0.9834\). Then \(\angle L=\arcsin(0.9834)\approx79.6^{\circ}\) or \(180 - 79.6 = 100.4^{\circ}\). But we need to check if these angles are valid. Let's check the sum of angles. For \(\angle L = 79.6^{\circ}\), \(\angle M=180 - 70 - 79.6 = 30.4^{\circ}\) (valid). For \(\angle L = 100.4^{\circ}\), \(\angle M=180 - 70 - 100.4 = 9.6^{\circ}\) (valid). But wait, the problem says "Find all possible values of \(\angle L\)". But maybe I made a mistake. Wait, actually, when \(k < l\) and \(\angle K\) is acute, and \(k>l\sin K\), then there are two possible triangles. Here, \(l\sin K\approx8.457\), \(k = 8.6\), so \(k>l\sin K\), so two possible triangles. But the problem has an option "Additional Angle" or "No Possible Angles". Wait, maybe I miscalculated. Wait, let's recalculate \(\sin L\):

\(\sin L=\frac{l\sin K}{k}=\frac{9\sin70^{\circ}}{8.6}\)

\(\sin70^{\circ}\approx0.9396926\)

\(9\times0.9396926 = 8.4572334\)

\(8.4572334\div8.6\approx0.9834\)

\(\arcsin(0.9834)\approx79.6^{\circ}\) (using calculator: \(\arcsin(0.9834)\) is approximately \(79.6\) degrees). The other angle is \(180 - 79.6 = 100.4\) degrees. But we need to check if \(l>k\), so \(\angle L>\angle K\) (since in a triangle, larger side is opposite larger angle). \(\angle K = 70^{\circ}\), so both \(79.6^{\circ}\) and \(100.4^{\circ}\) are greater than \(70^{\circ}\), so both are possible? But the problem has a button "Additional Angle" or "No Possible Angles". Wait, maybe I made a mistake in the Law of Sines. Wait, Law of Sines: \(\frac{\sin L}{l}=\frac{\sin K}{k}\), so \(\sin L=\frac{l\sin K}{k}\). Wait, \(l = 9\), \(k = 8.6\), so \(l>k\), so \(\angle L>\angle K\). \(\angle K = 70^{\circ}\), so \(\angle L\) must be greater than \(70^{\circ}\). Now, \(\sin L\approx0.9834\), so the reference angle is \(79.6^{\circ}\), and the other angle is \(180 - 79.6 = 100.4^{\circ}\). Both are greater than \(70^{\circ}\), so there are two possible angles? But let's check with the values. Wait, if \(\angle L = 100.4^{\circ}\), then \(\angle M=9.6^{\circ}\), and the sides: \(m\) can be calculated, but the problem is about angles. Wait, maybe the system considers that when \(k < l\) and \(\angle K\) is acute, and \(k>l\sin K\), then there are two possible angles. So the possible angles are \(79.6^{\circ}\) and \(100.4^{\circ}\)? But the problem says "to the nearest 10th of a degree". Wait, maybe I made a mistake in the calculation. Wait, let's use a calculator for \(\arcsin(0.9834)\). Let's calculate \(\arcsin(0.9834)\):

Using a calculator, \(\arcsin(0.9834)\approx79.6^{\circ}\) (since \(\sin80^{\circ}\approx0.9848\), so \(79.6^{\circ}\) is close). Then \(180 - 79.6 = 100.4^{\circ}\). So there are two possible angles. But the problem has an "Additional Angle" option, so maybe the first angle is \(79.6^{\circ}\) and the additional angle is \(100.4^{\circ}\). But let's check again. Wait, maybe the initial assumption is wrong. Wait, \(l = 9\), \(k = 8.6\), so side \(l\) is opposite \(\angle L\), side \(k\) is opposite \(\angle K\). So \(\frac{\sin L}{9}=\frac{\sin70^{\circ}}{8.6}\), so \(\sin L=\frac{9\sin70^{\circ}}{8.6}\approx0.9834\). Since \(\sin L\approx0.9834<1\), there are two possible angles: \(L_1=\arcsin(0.9834)\approx79.6^{\circ}\) and \(L_2 = 180^{\circ}-\arcsin(0.9834)\approx100.4^{\circ}\). Both angles are valid because \(L_1 + K=79.6 + 70=149.6<180\) and \(L_2+K = 100.4+70 = 170.4<180\), so the third angle in each case is positive. So there are two possible angles.

Answer:

First possible angle: \(79.6^{\circ}\), Additional angle: \(100.4^{\circ}\)