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

1 answer

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 \( \sin B \):

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

  1. 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 \( 14.5 , \text{cm} \).