Question

For each triangle shown below, determine whether you would use the Law of Sines or Law of Cosines to find angle x, and explain how you know which Law to use. Then find angle x to the nearest tenth.
NOTE: The perimeter of ABC-31

Answer 1

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 $3b-4,2b+1,b+10$ respectively.

The perimeter of the triangle ABC is 31.

$AB+BC+AC=31$

$(3b-4)+(2b+1)+(b+10)=31$

$6b+7=31$

Subtract 7 from both sides.

$6b=31-7$

$6b=24$

$b=\dfrac{24}{6}$

$b=4$

Now, the measures of the sides are:

$AB=3b-4$

$AB=3(4)-4$

$AB=12-4$

$AB=8$

$BC=2b+1$

$BC=2(4)+1$

$BC=8+1$

$BC=9$

And,

$AC=b+10$

$AC=4+10$

$AC=14$

Using the law of cosines, we get

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

$\cos A=\dfrac{(AC)^2+(AB)^2-(BC)^2}{2(AC)(AB)}$

$\cos A=\dfrac{(14)^2+(8)^2-(9)^2}{2(14)(8)}$

$\cos A=\dfrac{179}{224}$

Using calculator, we get

$\cos A=0.7991$

$A=\cos ^{-1}(0.7991)$

$x=36.9558^\circ$

$x\approx 37.0^\circ$

Therefore, the value of x is 37.0 degrees.