Question

using the law of sines for the asa case
determine the measures of all unknown angles and side lengths of △jkl. round side lengths to the nearest hundredth.
m∠k = 42
jk ≈
lk ≈ 7.11, 9.28, 10.52
(diagram: triangle jkl with ∠l = 28°, ∠j = 110°, side lj = 15)

Explanation:

Step1: Find the third angle

In a triangle, the sum of angles is \(180^\circ\). Given \(m\angle J = 110^\circ\), \(m\angle L = 28^\circ\), so \(m\angle K=180 - 110 - 28 = 42^\circ\) (matches the given value).

Step2: Apply Law of Sines for \(JK\)

Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let's denote:

• \(JL = 15\), opposite \(\angle K = 42^\circ\)
• \(JK\), opposite \(\angle L = 28^\circ\)

So \(\frac{JK}{\sin 28^\circ}=\frac{JL}{\sin 42^\circ}\)

Step3: Solve for \(JK\)

\(JK=\frac{JL\times\sin 28^\circ}{\sin 42^\circ}\)

\(JL = 15\), \(\sin 28^\circ\approx0.4695\), \(\sin 42^\circ\approx0.6691\)

\(JK=\frac{15\times0.4695}{0.6691}\approx\frac{7.0425}{0.6691}\approx10.52\)

Wait, but let's check for \(LK\) (maybe I misread the side). Wait, \(JL = 15\) (side \(JL\) is between \(J\) and \(L\), so side opposite \(\angle K\) is \(JL = 15\)? Wait, no: in \(\triangle JKL\), vertices are \(J\), \(K\), \(L\). So:

• Side \(JK\) is between \(J\) and \(K\), opposite \(\angle L = 28^\circ\)
• Side \(JL\) is between \(J\) and \(L\), length 15, opposite \(\angle K = 42^\circ\)
• Side \(LK\) is between \(L\) and \(K\), opposite \(\angle J = 110^\circ\)

Wait, maybe I messed up. Let's re - assign:

Let \(a = LK\) (opposite \(\angle J = 110^\circ\)), \(b = JK\) (opposite \(\angle L = 28^\circ\)), \(c = JL = 15\) (opposite \(\angle K = 42^\circ\))

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

For \(LK\) (a): \(\frac{a}{\sin 110^\circ}=\frac{15}{\sin 42^\circ}\)

\(a=\frac{15\times\sin 110^\circ}{\sin 42^\circ}\)

\(\sin 110^\circ=\sin(70^\circ)\approx0.9397\), \(\sin 42^\circ\approx0.6691\)

\(a=\frac{15\times0.9397}{0.6691}\approx\frac{14.0955}{0.6691}\approx21.07\) (not in options). Wait, the options are for \(JK\) or \(LK\)? The options given are 7.11, 9.28, 10.52. Let's check \(JK\) again.

\(JK\) is opposite \(\angle L = 28^\circ\), \(JL = 15\) opposite \(\angle K = 42^\circ\)

\(JK=\frac{15\times\sin 28^\circ}{\sin 42^\circ}\)

\(\sin 28^\circ\approx0.4695\), \(\sin 42^\circ\approx0.6691\)

\(15\times0.4695 = 7.0425\)

\(7.0425\div0.6691\approx10.52\)

Wait, but maybe the side \(JL = 15\) is opposite \(\angle K = 42^\circ\), and we need to find \(JK\) (opposite \(\angle L = 28^\circ\)) or \(LK\) (opposite \(\angle J = 110^\circ\)). But the options are 7.11, 9.28, 10.52. Let's recalculate \(JK\):

\(\frac{JK}{\sin 28^\circ}=\frac{JL}{\sin 42^\circ}\)

\(JK=\frac{15\times\sin 28^\circ}{\sin 42^\circ}\)

\(\sin 28^\circ\approx0.46947\), \(\sin 42^\circ\approx0.66913\)

\(15\times0.46947 = 7.04205\)

\(7.04205\div0.66913\approx10.52\)

So \(JK\approx10.52\)

Answer:

\(JK\approx10.52\)