Question

find the third side or each triangle to 1 decimal place.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula for finding side $AC$ (let's call it $a$) in $\triangle ABC$ is $a^{2}=b^{2}+c^{2}-2bc\cos A$. Here, $b = 11$, $c = 10$, and $A=109^{\circ}$.

Step2: Substitute values

$a^{2}=11^{2}+10^{2}-2\times11\times10\times\cos(109^{\circ})$. First, calculate the squares and the cosine - related part. $11^{2}=121$, $10^{2}=100$, and $\cos(109^{\circ})\approx - 0.326$. Then $2\times11\times10\times\cos(109^{\circ})\approx2\times11\times10\times(- 0.326)=-71.72$. So $a^{2}=121 + 100-(-71.72)=121+100 + 71.72=292.72$.

Step3: Find the value of $a$

$a=\sqrt{292.72}\approx17.1$.

Answer:

$17.1$