# How do you find the area of the triangle ABC with a = 7.2, c = 4.1, and B=28 degrees?

Nov 17, 2015

The area of the triangle is approximately $6.93$.

#### Explanation:

For this problem, we will use the following:
The law of cosines:
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right)$

Heron's formula:
$A r e a = \sqrt{p \left(p - a\right) \left(p - b\right) \left(p - c\right)}$ where $p = \frac{a + b + c}{2}$

Applying the law of cosines gives us

${b}^{2} = {a}^{2} + {c}^{2} - 2 a \mathcal{o} s \left(B\right)$

$\implies {b}^{2} = {7.2}^{2} + {4.1}^{2} - 2 \left(7.2\right) \left(4.1\right) \cos \left({28}^{\circ}\right)$

$\implies b = \sqrt{51.84 + 16.81 - 2 \left(7.2\right) \left(4.1\right) \cos \left({28}^{\circ}\right)} \approx 4.065$
(Note that for greater accuracy, we can perform the calculation of the square root at the very end, however for convenience we will use this approximation here)

Now, let $p = \frac{a + b + c}{2} \approx 7.6825$

By Heron's formula, then, we get the area $A$ of the triangle as

$A = \sqrt{p \left(p - 7.2\right) \left(p - 4.065\right) \left(p - 4.1\right)}$

Plugging this in gives us the final result

$A \approx 6.93$