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?

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Answer 1 · 2016-03-14T16:16:22

6.762 (6.762395692966 to be precise)

Answer can be verified here

Please use the above link to refer the image.

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 = $1/2 (base)(height) = 1/2 (1.7067)(8)$ =6.762

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