A triangle has sides A, B, and C. Sides A and B have lengths of 8 and 15, respectively. The angle between A and C is #(7pi)/24# and the angle between B and C is # (5pi)/8#. What is the area of the triangle?

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Answer 1 · 2016-11-29T17:18:45

If #alpha, beta# and #gamma# are tha angles opposite to #A,B# and #C#, the theorem of sines states that:

#fracA sinalpha = frac B sin beta = frac C sin gamma#

We know that:

#alpha = (5pi)/8#

#beta = (7pi)/24#

and as the sum of the internal angles of a triangle always equals #pi#

#gamma = pi-alpha-beta=pi/12#

We can use this to determine #C#:

#C = frac A sinalpha sin gamma#

Then we can use Eron's formula to calculate the area from the sides:

#S= sqrt(p(p-A)(p-B)(p-C))#

where #p=frac (A+B+C) 2#

As the angles are not such as the sine is immediately known we can use approximate values or do a bit of computation using trigonometric formulas.

#sin alpha = sin((5pi)/8) = sin (1/2*(5pi)/4) = sqrt((1-cos((5pi)/4))/2)=sqrt((1+cos((pi)/4))/2)=sqrt((1+sqrt(2)/2)/2)=sqrt(2+sqrt(2))/2#

#sin gamma = sin(pi/12) = ((sqrt(3)-1)/(2sqrt(2)))#

Then:

#A=8#
#B=15#
#C=8* frac cancel 2 (sqrt(2+sqrt(2))) * ((sqrt(3)-1)/(cancel 2sqrt(2)))#