Question

solve for c.
142°
c = ?
42
19°
c
round your final answer
to the nearest tenth.
law of sines: (\frac{a}{sin a} = \frac{b}{sin b} = \frac{c}{sin c})

Explanation:

Step1: Find the third angle

First, we know that the sum of the interior angles of a triangle is \(180^\circ\). Let the angles be \(A = 19^\circ\), \(B = 142^\circ\), and \(C\) (the third angle). So we calculate the third angle:
\(180 - 19 - 142 = 19^\circ\)? Wait, no, wait. Wait, the side opposite angle \(A\) (19°) is 42? Wait, no, let's label the triangle properly. Let's denote: angle at the left is \(19^\circ\) (let's call this angle \(A\)), the angle at the top is \(142^\circ\) (angle \(B\)), so the third angle (angle \(C\)) is \(180 - 19 - 142 = 19^\circ\)? Wait, no, that can't be. Wait, maybe I mixed up the sides. Wait, the side given is 42, which is opposite the 19° angle? Wait, no, let's re - label. Let's use the Law of Sines formula \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let's define: let angle \(A = 19^\circ\), angle \(B = 142^\circ\), then angle \(C=180-(19 + 142)=19^\circ\). Wait, so two angles are 19°, so it's an isosceles triangle? Wait, but the side opposite angle \(A\) (19°) is 42? Wait, no, the side with length 42 is opposite the 19° angle? Wait, no, the side we need to find is \(c\), which is opposite the 142° angle? Wait, no, let's do it correctly.

Let's denote:

- Angle \(A = 19^\circ\), opposite side \(a = 42\) (wait, no, the side with length 42 is adjacent to the 142° angle? Wait, looking at the diagram, the side of length 42 is between the 142° angle and the 19° angle? Wait, maybe I got the angles and sides mixed up. Let's start over.

The triangle has angles: 19°, 142°, and the third angle. Let's calculate the third angle: \(180-19 - 142=19^\circ\). Wait, so two angles are 19°, so the sides opposite them are equal? Wait, no, the side of length 42: which angle is opposite to it? Let's see, the side of length 42 is opposite the angle of \(19^\circ\) (the angle at the bottom left? No, the angle at the bottom left is 19°, and the side opposite to it would be the side opposite, which is the side with length... Wait, maybe the side of length 42 is opposite the 19° angle, and the side \(c\) is opposite the 142° angle.

So using the Law of Sines: \(\frac{c}{\sin B}=\frac{42}{\sin A}\), where \(A = 19^\circ\), \(B = 142^\circ\), and \(c\) is the side opposite angle \(B\).

First, confirm the angles: sum of angles in a triangle is \(180^\circ\). So \(19+142 + \text{third angle}=180\), so third angle \(=180-(19 + 142)=19^\circ\). Wait, so angle \(A = 19^\circ\), angle \(C = 19^\circ\), angle \(B = 142^\circ\). Then the side opposite angle \(A\) (19°) is 42, the side opposite angle \(C\) (19°) is also 42? But that contradicts the diagram. Wait, maybe I mis - identified the sides. Let's look at the diagram again: the triangle has a vertex angle of 142°, a base angle of 19°, and the side adjacent to the 142° and 19° angles is 42, and the side we need to find is the base \(c\).

Wait, let's use the Law of Sines correctly. Let's define:

- Let angle \(A = 19^\circ\), opposite side \(a = c\) (no, that's not right). Wait, maybe the side of length 42 is opposite the 19° angle, and the side \(c\) is opposite the 142° angle.

So according to the Law of Sines \(\frac{c}{\sin(142^\circ)}=\frac{42}{\sin(19^\circ)}\)

Now, we can solve for \(c\):

\(c=\frac{42\times\sin(142^\circ)}{\sin(19^\circ)}\)

Step2: Calculate the sines

First, calculate \(\sin(142^\circ)\) and \(\sin(19^\circ)\)

\(\sin(142^\circ)=\sin(180 - 38)^\circ=\sin(38^\circ)\approx0.6157\)

\(\sin(19^\circ)\approx0.3256\)

Step3: Compute \(c\)

Now substitute into the formula for \(c\):

\(c=\frac{42\times0.6157}{0.3256}\)

First, calculate the numerator: \(42\times0.6157 = 25.8594\)

Then divide by 0.3256: \(\frac{25.8594}{0.3256}\approx79.4\)

Answer:

\(79.4\)