A triangle has sides A, B, and C. The angle between sides A and B is pi/6 and the angle between sides B and C is pi/12. If side B has a length of 9, what is the area of the triangle?

1 Answer
Roella W.
Jan 27, 2016

A = 1/2B*B*(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)

Explanation:

The area of a triangle is given by A=1/2Bh where B is the base and h is the perpendicular height.
![Sketch - not accurate but is useful for the http://labels.](https://useruploads.socratic.org/Hh12S3h7RYeROH8qtYJC_triangle9.jpg)

We also know that the sum of the angles in a triangle is 180^o = pi

The angle between B and C is therefore (pi - pi/12 - pi/6)
=(12pi-pi-2pi)/12) = 9pi/12 =(3pi)/4

If h intersects with B so that x +y = B then
h/(B-y) = tan(pi/12) and h/y = tan(3/4pi)

Then y=h/tan(3/4pi)
:. h = (B-h/tan(3/4pi))tan(pi/12)

h = Btan(pi/12) - htan(pi/12)/tan(3/4pi)

So h(1+tan(pi/12)/tan(3/4pi)) = Btan(pi/12)

h((tan(3/4pi) + tan(pi/12))/tan(3/4pi)) = Btan(pi/12)

:. h = B(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)

So the area A = 1/2B*B(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)

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