Question

In∆ABC,A=54.2°,B=71.5°,a=12.4cm.Find b

Answer 1

To find side \( b \) in triangle \( ABC \) using the given angles and side \( a \), we can use the Law of Sines. The Law of Sines states:

\[
\frac{a}{\sin A} = \frac{b}{\sin B}
\]

Given:
- \( A = 54.2^\circ \)
- \( B = 71.5^\circ \)
- \( a = 12.4 \, \text{cm} \)

First, we need to calculate angle \( C \):

\[
C = 180^\circ - A - B
\]
\[
C = 180^\circ - 54.2^\circ - 71.5^\circ
\]
\[
C = 180^\circ - 125.7^\circ
\]
\[
C = 54.3^\circ
\]

Now we can use the Law of Sines to find side \( b \):

\[
\frac{b}{\sin B} = \frac{a}{\sin A}
\]
\[
b = \frac{a \cdot \sin B}{\sin A}
\]

Now plug in the values:

1. Calculate \( \sin A \) and \( \sin B \):

\[
\sin A = \sin(54.2^\circ) \approx 0.8090
\]
\[
\sin B = \sin(71.5^\circ) \approx 0.9490
\]

2. Substitute back into the formula for \( b \):

\[
b = \frac{12.4 \cdot 0.9490}{0.8090}
\]

Now calculate the value:

\[
b \approx \frac{11.76}{0.809} \approx 14.5 \, \text{cm}
\]

So, the length of side \( b \) is approximately \( 14.5 \, \text{cm} \).