Question

A triangle has sides A,B, and C. If the angle between sides A and B is `(pi)/8`, the angle between sides B and C is `pi/6`, and the length of B is 13, what is the area of the triangle?

Answer 1

`text{area} approx 20.3798`

Explanation:

First of all, let's rewrite in standard notation. It really helps to do these if we always label the triangle consistently.

The triangle has sides `a,b,c` (small letters) where `b=13` with opposing angles `C=pi/8=22.5^circ`, `A=pi/6=30^circ`.

The remaining angle is

`B=pi - A-C= pi-pi/8-pi/6 = { 17pi }/24 = 127.5^circ`

The Law of Sines says

`b/sin B = c/sin C`

`c = { b sin C }/ sin B = {13 sin(22.5^circ) }/ sin(127.5^circ )`

Now the area is

`text{area} = 1/2 bc sin A = frac{ (13)^2 sin(30^circ) sin(22.5^circ) } {2 \ sin 127.5^circ }`

I'm tempted to work out an exact answer, because those sines are all expressible using the usual operations including square root. But it's late, and the calculator says

`text{area} approx 20.3798`

Answer 2

`color(maroon)(A_t == 20.37 " sq units"`

Explanation:

`b = 13, hat A = pi/6, hat C = pi/8, hat B = pi - pi/6 - pi/8 = (17pi)/24`

Applying the Law of sines,

`a = (b sin A) / sin B = (13 * sin (pi/6)) / sin ((17pi)/24) ~~ 8.19`

Formula for area of " Delta, A_t = (1/2) * a b * sin C#

`color(maroon)(A_t = (1/2) * 8.19 * 13 * sin ( pi/8) = 20.37 " sq units"`