Given:
There are given that the triangle ABC.
Where,
[tex]\begin{gathered} AB=3.3km \\ AC=4.5km \\ \angle A=100^{\circ} \end{gathered}[/tex]
Explanation:
According to the question:
We need to find the value for BC:
So,
To find the value of BC, we need to use the cosine rule:
From the cosine rule:
[tex]BC^2=b^2+c^2-2bccosA[/tex]
Then,
Put the all values into the given formula:
[tex]\begin{gathered} BC^{2}=b^{2}+c^{2}-2bccosA \\ BC^2=(4.5)^2+(3.3)^2-2(4.5)(3.3)cos100^{\circ} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} BC^2=(4.5)^2+(3.3)^2-2(4.5)(3.3)cos100^{\operatorname{\circ}} \\ BC^2=20.25+10.89-29.7cos100^{\operatorname{\circ}} \\ BC^2=20.25+10.89-29.7(-0.17) \end{gathered}[/tex]
Then,
[tex]\begin{gathered} BC^{2}=20.25+10.89-29.7(-0.17) \\ BC^2=20.25+10.89+5.049 \\ BC^2=36.189 \\ BC=6.016 \end{gathered}[/tex]
Final answer:
Hence, the value of BC is 6.016 km.
