Given:
There are given that the triangle ABC.
Where,
[tex]\begin{gathered} a=35, \\ b=34, \\ c=28. \end{gathered}[/tex]Explanation:
To find the value of angles, we need to use the cosine rule:
So,
From the cosine rule:
[tex]\begin{gathered} a^2=b^2+c^2-2bccosA \\ b^2=a^2+c^2-2accosB \end{gathered}[/tex]Now,
First, find the angle for A,
So,
From the formula to find the angle A:
[tex]a^{2}=b^{2}+c^{2}-2bccosA[/tex]Then,
Put the value of a, b, and c into the above formula:
So,
[tex]\begin{gathered} a^{2}=b^{2}+c^{2}-2bccosA \\ 35^2=34^2+28^2-2\times34\times28\times cosA \\ 1225=1156+784-1904cosA \\ 1225=1940-1904cosA \end{gathered}[/tex]Then,
Subtract 1940 from both side of the equation:
[tex]\begin{gathered} 1225=1940-1904cosA \\ 1225-1940=1940-1,904cosA-1940 \\ -715=-1904cosA \end{gathered}[/tex]Then,
Divide by -1904 in both side of the above equation:
So,
[tex]\begin{gathered} -715=-1904cosA \\ -\frac{715}{-1904}=-\frac{1904}{-1904}cosA \\ 0.3755=cosA \\ A=cos^{-1}0.3755 \\ A=67.9 \end{gathered}[/tex]Now,
From the formula to find the angle for B:
[tex]\begin{gathered} b^{2}=a^{2}+c^{2}-2accosB \\ 34^2=35^2+28^2-2\times35\times28\times cosB \\ 1156=1225+784-1960cosB \\ 1156=2009-1960cosB \end{gathered}[/tex]Then,
Subtract 2009 from both sides of the equation:
So,
[tex]\begin{gathered} 1156=2009-1960cosB \\ 1156-2009=2009-1960cosB-2009 \\ -853=-1960cosB \\ cosB=\frac{853}{1960} \\ cosB=0.435 \\ B=cos^{-1}(0.435) \\ B=64.2^{\degree} \end{gathered}[/tex]Now,
To find the angle C, we need to use the interior angle concept:
From the interior angle concept, addition of interior angle of any triangle is equal to 180 degrees.
Then,
[tex]\angle A+\angle B+\angle C=180^{\degree}[/tex]Then,
Put the value of angle A and angle B into the above formula:
So,
[tex]\begin{gathered} \angle A+\angle B+\angle C=180^{\degree} \\ 67.9^{\degree}+64.2^{\degree}+\angle C=180^{\degree} \\ 132.1^{\degree}+\operatorname{\angle}C=180^{\operatorname{\degree}} \\ \operatorname{\angle}C=180^{\degree}-132.1^{\degree} \\ \angle C=47.9 \end{gathered}[/tex]Final answer:
Hence, the value of angles A, B, and C are shown below:
[tex]\begin{gathered} \angle A=67.9^{\degree} \\ \angle B=64.2^{\degree} \\ \angle C=47.9^{\degree} \end{gathered}[/tex]