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)

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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