The length of the side AC = 18.9 units and measure of ∠C = 68°
What is the sine rule of triangle?
"It states that the ratio of the side length of a triangle to the sine of the opposite angle, which is the same for all three sides."
For triangle PQR,
[tex]\frac{sin~P}{p} =\frac{sin~Q}{q}= \frac{sin~R}{r}[/tex]
What is the cosine rule of triangle?
For triangle PQR,
[tex]a^2 = b^2 + c^2 - 2bc~ cos \angle A\\b^2 = a^2 + c^2 - 2ac~ cos \angle B\\c^2 = a^2 + b^2 - 2ab~ cos \angle C\\[/tex]
where BC = a, AC = b, AB = c
For given question,
We have been given two sides and one angle of triangle ABC.
AB = 12, BC = 18, and m(∠B) = 75
Using cosine rule for triangle ABC,
[tex]\Rightarrow AC^2 = BC^2 + AB^2 - 2(BC)(AB)~ cos \angle B\\\\\Rightarrow AC^2 = 18^2 + 12^2 - 2(18)(12)~ cos(75)\\\\\Rightarrow AC^2=324+144-(432\times 0.2588)\\\\\Rightarrow AC^2=468-111.81\\\\\Rightarrow AC^2=356.19\\\\\Rightarrow AC=18.87\\\\\Rightarrow AC\approx 18.9[/tex]
Using sine rule for triangle ABC,
[tex]\frac{sin~A}{BC} =\frac{sin~B}{AC}= \frac{sin~C}{AB}[/tex]
Consider,
[tex]\Rightarrow \frac{sin~A}{BC} =\frac{sin~B}{AC}\\\\\Rightarrow \frac{sin~A}{18} =\frac{sin~75}{18.9}\\\\\Rightarrow 18.9\times sin(A)=18\times sin(75)\\\\\Rightarrow 18.9\times sin(A)=18\times 0.97\\\\\Rightarrow sin~A=0.9238\\\\\Rightarrow \angle C=sin^{-1}(0.9238)\\\\\Rightarrow \angle C = 67.5^{\circ}\\\\\Rightarrow \angle C=68^{\circ}[/tex]
Therefore, the length of the side AC = 18.9 units and measure of ∠C = 68°
Learn more about the sine and cosine rule for triangle here:
https://brainly.com/question/20839703
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