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 15, what is the area of the triangle?

Drawing a clear diagram is essential for this kind of problem. Mine is not perfectly to scale, but it allows me to have clear in my own mind what information I have and what I'm trying to achieve.

(I've placed the image at the bottom of the page because the formatting works better that way.)

We know that the area of a triangle is given by:

#Area=1/2bh#

We already know that the base is #15# #cm#, but we don't know the perpendicular height. I have constructed a right angled triangle, so we could use trig, but we don't know how far along the #15# #cm# side the added line hits.

We could use sine rule or cosine rule in the main triangle (they work in all triangles, not just right-angled ones), but we still need a little more information.

In radians, the angles in a triangle add up to #pi# (or #180^o#, if we worked in degrees). We already have #angleAB#=#pi/4=(3pi)/12# and #angleBC#=#pi/12# for a total of #(4pi)/12#, so #angleAC# (all of it) must be #(8pi)/12#.

Now we can use the sine rule to find the length of #A# or #C# or both:

#C/sinc=B/sinc to C = Bsinc/sinb = 15(sin(pi/4))/sin((8pi)/12) = 12.2# #cm#

(for our purposes here, lowercase letters represent angles and uppercase represent sides, though the convention is usually the other way around)

Now that we know the length of #C#, we can use trig to find the height of the vertical line: the height of the triangle. Let's call that #'D'#:

#cos(pi/12)=D/C to D=Ccos(pi/12)=12.2cos(pi/12)=11.78# #cm#

Now we know the base and height of the triangle, we can calculate its area:

#Area=1/2bh=1/2*15*11.78=88.35# #cm^2#