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.