Question

Use the Law of Cosines to find all the interior angles of a triangle having sides that
measure 12.6 feet, 7.4 feet, and 8.2 feet. Use Heron's formula to find the area of the triangle.

Answer 1

Task. Use the Law of Cosines to find all the interior angles of a triangle having sides that measure 12.6 feet, 7.4 feet, and 8.2 feet. Use Heron’s formula to find the area of the triangle.

Solution. Let $a=12.6$, $b=7.4$ and $c=8.2$ be the sides of the triangle, and $\alpha$, $\beta$, $\gamma$ are the opposite angles.

Then the Law of Cosines claims that

$c^{2}=a^{2}+b^{2}-2ab\cos\gamma,$

whence

$\gamma=\arccos\frac{a^{2}+b^{2}-c^{2}}{2ab}.$

Similarly,

$\beta=\arccos\frac{a^{2}+c^{2}-b^{2}}{2ac}.$

$\alpha=\arccos\frac{b^{2}+c^{2}-a^{2}}{2bc}.$

Thus

$\alpha=\arccos\frac{b^{2}+c^{2}-a^{2}}{2bc}=\arccos\frac{7.4^{2}+8.2^{2}-12.6^{2}}{2*7.4*8.2}=\arccos\frac{-36.760}{121.36}$

$=\arccos(-0.30290)=107.632{}^{\circ}$

$\beta=\arccos\frac{a^{2}+c^{2}-b^{2}}{2ac}=\arccos\frac{12.6^{2}+8.2^{2}-7.4^{2}}{2*12.6*8.2}=\arccos\frac{171.24}{206.64}$

$=\arccos(0.82869)=34.036{}^{\circ}$

$\gamma=\arccos\frac{a^{2}+b^{2}-c^{2}}{2ab}=\arccos\frac{12.6^{2}+7.4^{2}-8.2^{2}}{2*12.6*7.4}=\arccos\frac{146.28}{186.48}$

$=\arccos(0.78443)=38.332{}^{\circ}$

Now let us find the area of the triangle using Heron’s formula:

$S=\sqrt{p(p-a)(p-b)(p-c)},$

where

$p=\frac{a+b+c}{2}$

is the half-perimeter. Thus

$p=\frac{12.6+7.4+8.2}{2}\frac{28.2}{2}=14.1\text{ feet}.$

Then

$S$ $=\sqrt{14.1*(14.1-12.6)*(14.1-7.4)*(14.1-8.2)}$

$=\sqrt{14.1*1.5*6.7*5.9}=\sqrt{863.06}\approx 28.915\text{ feet}^{2}.$