A triangle has sides A, B, and C. Sides A and B have lengths of 5 and 3, respectively. The angle between A and C is (19pi)/24 and the angle between B and C is  (pi)/8. What is the area of the triangle?

Sep 28, 2016

$A \approx 1.94 u n i t {s}^{2}$

Explanation:

Let's use the standard notation where the lengths of the sides are the lowercase letters, a, b, and c and the angles opposite the sides are the corresponding uppercase letters, A, B, and C.

We are given $a = 5 , b = 3 , A = \frac{19 \pi}{24} , \mathmr{and} B = \frac{\pi}{8}$

We can compute angle C:

$\frac{24 \pi}{24} - \frac{19 \pi}{24} - \frac{3 \pi}{24} = \frac{2 \pi}{24} = \frac{\pi}{12}$

We can compute the length of side c using either the law of sines or the law of cosines. Let's use the law of cosines, because it does not have the ambiguous case problem that the law of sines has:

c² = a² + b² - 2(a)(b)cos(C)

c² = 5² + 3² - 2(5)(3)cos(pi/12)

$c = \sqrt{5.02}$

Now we can use Heron's Formula to compute the area:

Correction made to the following lines:

$p = \frac{5 + 3 + \sqrt{5.02}}{2} \approx 5.12$

A = sqrt(5.12(5.12 - 5)(5.122 - 3)(5.12 - sqrt5.02)

$A \approx 1.94$
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