Question

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?

Answer 1

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*A*B*cos(c)`
or
`C=sqrt(A^2+B^2-2*A*B*cos(c))`

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`

`Area=3.75`