A triangle has sides A,B, and C. If the angle between sides A and B is $(5pi)/8$ , the angle between sides B and C is $pi/4$ , and the length of B is 19, what is the area of the triangle?

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Answer 1 · 2016-01-14T12:29:10

Area $=1/2*19*32.44 ~~308.13$

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The area of a triangle $=1/b*h$ where $b =$ base and $h=$ height

In this case $tan((5pi)/8) =h/x$ and $tan(pi/4) = h/y$ where

$x+y = B =19$
$y=19-x$

So $xtan((5pi)/8) =ytan(pi/4)$
$xtan((5pi)/8) = (19-x)tan(pi/4)$
$x(tan((5pi)/8) +tan(pi/4)) = 19tan(pi/4)$

$:.x = 19tan(pi/4)/(tan((5pi)/8)+tan(pi/4))$

$h = (19tan(pi/4)/(tan((5pi)/8)+tan(pi/4)))tan((5pi)/8)$
$~~(19*1*2.414)/(2.414+1)$
$~~32.44$

Area $=1/2*19*32.44 ~~308.13$

A triangle has sides A,B, and C. If the angle between sides A and B is (5pi)/8(5pi)/8, the angle between sides B and C is pi/4pi/4, and the length of B is 19, what is the area of the triangle?