Question

A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is `(13pi)/24` and the angle between B and C is `(pi)24`. What is the area of the triangle?

Answer 1

Since triangle angles add to `pi` we can figure out the angle between the given sides and the area formula gives
`A = \frac 1 2 a b sin C = 10(sqrt{2} + sqrt{6})`.

Explanation:

It helps if we all stick to the convention of small letter sides `a,b,c` and capital letter opposing vertices `A,B,C`. Let's do that here.

The area of a triangle is `A = 1/2 a b sin C` where `C` is the angle between `a` and `b`.

We have `B=\frac{ 13\pi}{24 }` and (guessing it's a typo in the question) `A=\pi/24`.

Since triangle angles add up to `180^\circ` aka `\pi` we get

`C = \pi - \pi/24 - frac{13 pi}{24} = \frac{10 pi}{24} =\frac{5pi}{12}`

`\frac{5pi}{12}` is `75^\circ.` We get its sine with the sum angle formula:

`sin 75^circ = sin(30 + 45) = sin 30 cos 45 + cos 30 sin 45`

`= (\frac 1 2 + frac sqrt{3} 2) \sqrt{2}/ 2`

`= \frac 1 4(sqrt(2) + sqrt(6))`

So our area is

`A = \frac 1 2 a b sin C = \frac 1 2 (10)(8) \frac 1 4(sqrt(2) + sqrt(6))`

`A = 10(sqrt{2} + sqrt{6})`

Take the exact answer with a grain of salt because it's not clear we guessed correctly what the asker meant by the angle between `B` and `C`.