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:

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

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(7pi)/24 and the angle between B and C is (5pi)/8 (5pi)/8. What is the area of the triangle?