Given:
The figure of a triangle.
The perimeter of the triangle ABC is 31.
To find:
The value of x in the given triangle.
Solution:
Three sides of the triangle ABC are AB, BC, AC are their measures are [tex]3b-4,2b+1,b+10[/tex] respectively.
The perimeter of the triangle ABC is 31.
[tex]AB+BC+AC=31[/tex]
[tex](3b-4)+(2b+1)+(b+10)=31[/tex]
[tex]6b+7=31[/tex]
Subtract 7 from both sides.
[tex]6b=31-7[/tex]
[tex]6b=24[/tex]
[tex]b=\dfrac{24}{6}[/tex]
[tex]b=4[/tex]
Now, the measures of the sides are:
[tex]AB=3b-4[/tex]
[tex]AB=3(4)-4[/tex]
[tex]AB=12-4[/tex]
[tex]AB=8[/tex]
[tex]BC=2b+1[/tex]
[tex]BC=2(4)+1[/tex]
[tex]BC=8+1[/tex]
[tex]BC=9[/tex]
And,
[tex]AC=b+10[/tex]
[tex]AC=4+10[/tex]
[tex]AC=14[/tex]
Using the law of cosines, we get
[tex]\cos A=\dfrac{b^2+c^2-a^2}{2bc}[/tex]
[tex]\cos A=\dfrac{(AC)^2+(AB)^2-(BC)^2}{2(AC)(AB)}[/tex]
[tex]\cos A=\dfrac{(14)^2+(8)^2-(9)^2}{2(14)(8)}[/tex]
[tex]\cos A=\dfrac{179}{224}[/tex]
Using calculator, we get
[tex]\cos A=0.7991[/tex]
[tex]A=\cos ^{-1}(0.7991)[/tex]
[tex]x=36.9558^\circ[/tex]
[tex]x\approx 37.0^\circ[/tex]
Therefore, the value of x is 37.0 degrees.
