# A triangle has sides A, B, and C. The angle between sides A and B is pi/4 and the angle between sides B and C is pi/12. If side B has a length of 8, what is the area of the triangle?

Mar 14, 2016

6.762 (6.762395692966 to be precise)

#### Explanation:

Now that we have the triangle (blue coloured in the above link), let's begin.

First drop a perpendicular from vertex A to the side BC, let it meet BC at the point D.

Now, assume the length BD = $x$
Thus, CD = $8 - x$
(since BC = $8$)

using trigonometry, we have:

from triangle ADB, AD= x cos(45º) (because angle B=45º)
from triangle CDB, AD=(8-x)cos(15º) (because angle C=π/12=15º)

hence we have,
x cos(45º)=(8-x)cos(15º)
solve this linear equation in x to get the value of x as 2.4136.

hence AD = x cos(45º) = 1.7067

Now, looking again at the triangle ABC whose area is to be found:
We have the height (AD=1.7067), we have the base (BC=8).

Thus, area of the triangle = $\frac{1}{2} \left(b a s e\right) \left(h e i g h t\right) = \frac{1}{2} \left(1.7067\right) \left(8\right)$=6.762