A triangle has sides A, B, and C. The angle between sides A and B is #(5pi)/12# and the angle between sides B and C is #pi/6#. If side B has a length of 17, what is the area of the triangle?

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Answer 1 · 2016-12-29T18:18:28

The area is 144.5.

let's first switch to standard notation. Sides are denotes a, b and c and the angle of the vertex opposite the sides are denotes A, B and C.

The question then becomes, what is the area of the triangle where #b=17#, #A=pi/6# and #C=(5pi)/12#.

The best equation for the areas in #Area=1/2ab sinC#. For this we need the length of side a. To calculate this we also, need angle B.

To calculate angle B we use the fact that #A+B+C=pi#. We rewrite this as #B=pi-A-C#

#B=pi - pi/6-(5pi)/12=(5pi)/12#

Using the sine rule:

#a/(sin A) = b/(sin B)#

This gives:

#a=(b sin A)/(sin B)= (17 sin (pi/6)) /( sin (5pi)/12)#

So the areas is:

#Area = 1/2 (b sin A)/(sin B)= (17 sin (pi/6)) /( sin (5pi)/12) 17 sin (5pi)/12 =17^2 sin(pi/6)#

Now #sin(pi/6)=1/2#

So:

#Area = 17^2/2=289/2=144.5#