# A triangle has sides A, B, and C. Sides A and B have lengths of 3 and 5, respectively. The angle between A and C is (13pi)/24 and the angle between B and C is  (7pi)/24. What is the area of the triangle?

Dec 31, 2015

By use of 3 laws:

• Sum of angles
• Law of cosines
• Heron's formula

The area is 3.75

#### Explanation:

The law of cosines for side C states:

${C}^{2} = {A}^{2} + {B}^{2} - 2 \cdot A \cdot B \cdot \cos \left(c\right)$
or
$C = \sqrt{{A}^{2} + {B}^{2} - 2 \cdot A \cdot B \cdot \cos \left(c\right)}$

where 'c' is the angle between sides A and B. This can be found by knowing that the sum of degrees of all angles is equal to 180 or, in this case speaking in rads, π:

a+b+c=π
c=π-b-c=π-13/24π-7/24π=24/24π-13/24π-7/24π=(24-13-7)/24π=4/24π=π/6
c=π/6

Now that the angle c is known, side C can be calculated:

C=sqrt(3^2+5^2-2*3*5*cos(π/6))=sqrt(9+25-30*sqrt(3)/2)=8.019

$C = 2.8318$

Heron's formula calculates the area of any triangle given the 3 sides by calculating half of the perimeter:

τ=(A+B+C)/2=(3+5+2.8318)/2=5.416

and using the formula:

Area=sqrt(τ(τ-A)(τ-B)(τ-C))=sqrt(5.416(5.416-3)(5.416-5)(5.416-2.8318))=3.75

$A r e a = 3.75$