How do you solve the triangle given $A = 116^{\circ}, b = 5, c = 3$ $A=116^circ, b=5, c=3$ ?

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Answer 1 · 2017-02-06T06:03:40

$a = 6.87$ $a=6.87$ ;
$hatB=40.88°$ ;
$hatC=23.12°$

By the cosine theorem, it is:

$a^2=b^2+c^2-2bc cos hatA$

Then

$a^2=5^2+3^2-2*5*3*cos116°$

$a^2=25+9-30*(-0.44)=47.15$

Then $a=sqrt(47.15)=6.87$

By the sines theorem, you get:

$a/sin hatA=b/sin hatB$

$->sin hatB=(b*sin hatA)/a$

$=(5*sin116°)/6.87=0.65$

then $hatB=sin^-1 0.65=40.88°$

$hatC=180°-hatA-hatB=180°-116°-40.88°=23.12°$

How do you solve the triangle given A=116^circ, b=5, c=3A=116∘,b=5,c=3A=116^circ, b=5, c=3?