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Precalculus Mathematics Homework Help
Find angle B in the trigonometry problem
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[QUOTE="chwala, post: 6527928, member: 287397"] Okay, i managed to do this, ##tan 80##=## \frac {AE}{ED}## and we know that; ##AE=x sin 40^0## therefore, ##DE##=## \dfrac {x sin 40^0⋅cos 80^0}{sin 80^0}## ##tan B##=## \dfrac {x⋅sin 40^0⋅sin 80^0}{x⋅sin 80^0+x⋅sin 40^0⋅cos 80^0}##.......(1) ##[cos 40^0-cos 120^0]=2sin 80^0sin 40^0##→##0.5[cos 40^0-cos 120^0]=sin 80^0sin 40^0## ##[sin 120^0-sin 40^0]= 2 cos 80^0 sin40^0##→##0.5[sin 120^0-sin 40^0]= cos 80^0 sin40^0## ##tan B##=## \dfrac {-0.5cos 120^0+0.5cos40^0}{sin 80^0+0.5sin 120^0-0.5sin40^0}## ##tan B##=## \dfrac {0.25+0.5 \dfrac {EC}{x}}{\dfrac {AE}{AD}+0.25 \sqrt 3-0.5 \dfrac {AE}{x}}## alternatively, ##cos 40^0##=## \dfrac {sin 80^0}{2 sin 40^0}## therefore from (1) above we shall have, ##tan B##=## \dfrac {sin 80^0}{ 2 cos 40^0 +cos 80^0}## on dividing each term by ##sin 80^0## ##tan B##=## \dfrac {1}{\frac {1}{sin 40^0}+\dfrac {cos 80^0}{sin 80^0}}##=## \dfrac {1}{\frac {1}{sin 40^0}+\dfrac {sin 10^0}{sin 80^0}}## Neil am stuck here mate :cool: [/QUOTE]
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Precalculus Mathematics Homework Help
Find angle B in the trigonometry problem
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