### Home > PC > Chapter 1 > Lesson 1.3.2 > Problem1-114

1-114.

Harold, the happy concrete guy, got a call on the phone about some concrete he has to pour for an odd shaped triangular patio. The only information he got was that the side lengths were $5$, $6$, and $8$ meters long.

1. Find the angles for Harold so he can stay happy.

Draw and label a diagram.

Since no angle measures are given, what formula should be used?

$6^2=5^2+8^2-2(5)(8)\cos{A}$

$36=89-80\cos{A}$

$-53 = -80\cos{A}$

$\frac{53}{80}=\cos{A}$

Law of Cosines

$a² = b² + c² - 2bc\left(\text{cos }A\right)$

Use the Law of Sines to find another angle.

$\frac{a}{\text{Sin}A} = \frac{b}{\text{Sin}B} = \frac{c}{\text{Sin}C}$

1. Find the area of the patio that Harold will be forming.

Draw a segment from angle C ⊥ to side $AB$.
Label the intersection point $E$.

Using the sine function, calculate the height, $EC$.

Determine the area of $ΔABC$ by calculating half of the base ($8$) times the height ($EC$).

$14.98 \text{ sq. meters}$