How do you solve the triangle given $A=55^circ, b=3, c=10$ ?

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Answer 1 · 2018-02-01T14:59:18

$a=8.64$ units long

I'm assuming that you are trying to find side $a$ of the triangle.

Well, the cosine rule states that

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

Plugging in $b=3$ , $c=10$ , $A=55^@$ , we get

$a^2=9+100-60 cos55^@$

$a^2=109-34.41$

$a^2=74.59$

$a=sqrt(74.59)~~8.64$

So, side $a$ will be $8.64$ units long.

How do you solve the triangle given A=55^circ, b=3, c=10A=55^circ, b=3, c=10?