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?

Question

1392 views around the world

You can reuse this answer
Creative Commons License

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:

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

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