Question

In the given problem solve the triangle using either the law of sines or law of cosines- $a=4,c=5,B=55°$

Short answer

Required values of the triangle are$b=4.25,A=50.45°,C=74.55°$

Step-by-step solution

Step 1. Given information

Given values are $a=4,c=5,B=55°$.

We know for a triangle with sides a, b, c and opposite angles A, B, C, respectively, the Law of Cosines are:

$$\begin{gathered} c^2=a^2+b^2-2ab\cos C \\ {b}^2=a^2+c^2-2ac\cos B \\ {a}^2=b^2+c^2-2bc\cos A \end{gathered}$$

Step 2. Calculation

$Forb,$

$$\begin{gathered} b^2=a^2+c^2-2ac\cos B \\ =4^2+5^2-2\times 4\times 5\times \cos \left(55°\right) \\ =16+25-40\times 0.5735 \\ =41-22.94 \\ =18.06 \\ b=4.25 \end{gathered}$$

$ForA,$

localid="1646578213238" $$\begin{gathered} \cos A=\raisebox{1ex}{$b^2+c^2-a^2$}\!\left/ \!\raisebox{-1ex}{$2bc$}\right. \\ =\raisebox{1ex}{${\left(4.25\right)}^2+5^2-4^2$}\!\left/ \!\raisebox{-1ex}{$2\times 4.25\times 5$}\right. \\ =\raisebox{1ex}{$18.0625+25-16$}\!\left/ \!\raisebox{-1ex}{$42.5$}\right. \\ =0.6367 \\ A={\cos }^{-1}\left(0.6367\right) \\ =50.45° \end{gathered}$$

$ForC,$

$$\begin{gathered} C=180°-A-B \\ =180°-50.45°-55° \\ =74.55° \end{gathered}$$