+105 so a triangle is 180 degress. i have a word problem and i can only seem to get as high as 176 degrees.
Problem
suppose you have a triangle where A=47(degrees), side b=7.00, and side c=3.59. Use cosine law to find side a, angle B and angle C.
my answers so far are..
a=5.25
B=99
C=30.34
=176.
any help would be greatly appricated!!!
+26396 Hi Strider,
α=47\ensuremath∘b=7c=3.59side a?angle β ? and angle γ ?
a:
a2=b2+c2−2bc∗cos(47\ensuremath∘)a2=72+3.592−2∗7∗3.59∗cos(47\ensuremath∘)a2=27.6108624233a=5.25460392639
β:
b2=a2+c2−2ac∗cos(β)cos(β)=a2+c2−b22accos(β)=−0.22532402766β=103.021932880\ensuremath∘
γ:
c2=a2+b2−2ab∗cos(γ)cos(γ)=a2+b2−c22abcos(γ)=0.86621674081γ=29.9780671201\ensuremath∘
α+β+γ=47\ensuremath∘+103.021932880\ensuremath∘+29.9780671201\ensuremath∘=180.000000000\ensuremath∘
+26396 Hi Strider,
α=47\ensuremath∘b=7c=3.59side a?angle β ? and angle γ ?
a:
a2=b2+c2−2bc∗cos(47\ensuremath∘)a2=72+3.592−2∗7∗3.59∗cos(47\ensuremath∘)a2=27.6108624233a=5.25460392639
β:
b2=a2+c2−2ac∗cos(β)cos(β)=a2+c2−b22accos(β)=−0.22532402766β=103.021932880\ensuremath∘
γ:
c2=a2+b2−2ab∗cos(γ)cos(γ)=a2+b2−c22abcos(γ)=0.86621674081γ=29.9780671201\ensuremath∘
α+β+γ=47\ensuremath∘+103.021932880\ensuremath∘+29.9780671201\ensuremath∘=180.000000000\ensuremath∘
