How do you solve the triangle given $C = 145^{\circ}, b = 4, c = 14$ $C=145^circ, b=4, c=14$ ?

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Answer 1 · 2017-02-08T06:28:24

$hatA~=25.57°$
$hatB~=9.43°$
$a~=10.56$

By the Euler Theorem, you know that:

$b/sin hatB=c/sin hatC$

and you can find

$sin hatB=(b sin hatC)/c$

$=(cancel4^2*sin 145°)/cancel14^7=(2*0.57)/7~=0.16$

and $hatB~=9.43°$

Then $hatA=180°-hatB-hatC$

$~=180°-145°-9.43°=25.57$

By the used theorem, it is:

$a/sin hatA=c/sin hatC$

and therefore is

$a=(c*sin hatA)/sin hatC$

$~=(14*0.43)/0.57=10.56$

How do you solve the triangle given C=145^circ, b=4, c=14C=145∘,b=4,c=14C=145^circ, b=4, c=14?