EXPLANATION:
Given;
We are given the following information for a triangle;
[tex]AC=7,BC=9,AB=4[/tex]Required;
We are required to find the measures of angles A, B and C.
Step-by-step solution;
For any triangle with three sides given/available we can apply the cosine rule to calculate the missing angle(s).
The cosine rule is quoted below;
[tex]\begin{gathered} a^2=b^2+c^2-2bcCosA \\ \\ OR \\ \\ b^2=a^2+c^2-2acCosB \\ \\ OR \\ \\ c^2=a^2+b^2-2abCosC \end{gathered}[/tex]We shall now take angle A to begin with;
We start by making Cos A the subject of the equation;
[tex]\begin{gathered} a^2=b^2+c^2-2bcCosA \\ Add\text{ }2bcCosA\text{ }to\text{ }both\text{ }sides: \\ \\ 2bcCosA+a^2=b^2+c^2 \\ \\ Subtract\text{ }a^{2\text{ }}from\text{ }both\text{ }sides: \\ \\ 2bcCosA=b^2+c^2-a^2 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }2bc: \\ \\ CosA=\frac{b^2+c^2-a^2}{2bc} \end{gathered}[/tex]With the refined form of the cosine formula we can now substitute and solve as follows;
[tex]CosA=\frac{7^2+4^2-9^2}{2(7)(4)}[/tex][tex]\begin{gathered} CosA=\frac{49+16-81}{56} \\ \\ CosA=\frac{-16}{56} \\ \\ CosA=-0.285714285714 \end{gathered}[/tex]We can now use a calculator to find the value of arccos -0.2857...
[tex]Cos^{-1}A=106.601549599[/tex]Rounded to the nearest degree the angle measure is;
[tex]\angle A=107\degree[/tex]To solve for angle B, we can now apply the sine rule, since we have two sides and one of the angles. The sine rule is quoted below;
[tex]\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}[/tex]We have angle A, and side length a. We have side length b, which means we can solve for angle B. This is shown below;
[tex]\begin{gathered} \frac{a}{SinA}=\frac{b}{SinB} \\ \\ Cross\text{ }multiply: \\ \\ SinB=\frac{b\times SinA}{a} \end{gathered}[/tex][tex]\begin{gathered} SinB=\frac{7\times Sin107\degree}{9} \\ \\ SinB=\frac{7\times0.956304755963}{9} \\ \\ SinB=0.743792587971 \end{gathered}[/tex]We can now use a calculator and find the arcsin of the decimal 0.7437...
[tex]Sin^{-1}B=48.0554959457[/tex]Rounded to the nearest degree, we now have;
[tex]\angle B=48\degree[/tex]Note that the sum of angles in a triangle all sum up to 180 degrees.
We have angles A and B. We can calculate angle C as follows;
[tex]\begin{gathered} \angle A+\angle B+\angle C=180 \\ \\ 107+48+\angle C=180 \\ \\ 155+\angle C=180 \end{gathered}[/tex]Subtract 155 from both sides;
[tex]\angle C=25\degree[/tex]Therefore, the angles are;
ANSWER:
[tex]\begin{gathered} \angle A=107\degree \\ \\ \angle B=48\degree \\ \\ \angle C=25\degree \end{gathered}[/tex]