Answer:
Side CA = 7.8
Step-by-step explanation:
Given:
Acute angled [tex]\triangle ABC[/tex].
[tex]\angle B =40^\circ[/tex]
AB = 10
BC = 12
We can use cosine rule here to find the side AC = b
Formula for cosine rule:
[tex]cos B = \dfrac{a^{2}+c^{2}-b^{2}}{2ac}[/tex]
Where
a is the side opposite to [tex]\angle A[/tex]
b is the side opposite to [tex]\angle B[/tex]
c is the side opposite to [tex]\angle C[/tex]
[tex]cos 40 = \dfrac{12^{2}+10^{2}-b^{2}}{2\times 12\times 10}\\\Rightarrow cos 40 = \dfrac{144+100-b^{2}}{240}\\\Rightarrow 0.77 = \dfrac{244-b^{2}}{240}\\\Rightarrow 244-b^{2} = 0.77 \times 240\\\Rightarrow 244-b^{2} = 183.85\\\Rightarrow 244-183.85 = b^{2}\\\Rightarrow b^2 = 60.15\\\Rightarrow b = 7.76[/tex]
To the nearest tenth b = 7.8
