# How do you find the area of the triangle given A=5^circ15', b=4.5, c=22?

May 9, 2017

Use ${A}_{\triangle} = \frac{1}{2} \left(a\right) \left(b\right) \sin \left(C\right)$.

#### Explanation:

One of the many formulas to find the area of a triangle is:
${A}_{\triangle} = \frac{1}{2} \left(a\right) \left(b\right) \sin \left(C\right)$ where $a$ and $b$ and side lengths and $C$ is the angle between sides $a$ and $b$

For this problem, we first convert angle A to degrees:
Since there are 60 minutes in a degree and $\frac{15}{60} = .25$, ${5}^{\circ} 15 ' = {5.25}^{\circ}$.

Now we simply plug the values into the area formula above:
${A}_{\triangle} = \frac{1}{2} \left(4.5\right) \left(22\right) \sin \left({5.25}^{\circ}\right)$
$= 4.529$ ${\text{units}}^{2}$

Therefore, the area of the triangle is 4.529 sq. units.