How do you solve the triangle given #triangleABC, a=22, b=18, mangleC=130#?

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Answer 1 · 2017-01-04T16:42:30

Solution is #a=22#, #b=18#, #c=36.29#,

#m/_A=27.67^@#, #m/_B=22.33^@# and #m/_C=130^@#

Solving a triangle means finding all sides and angles of the triangle. For this we use sine formulas i.e.

#a/sinA=B/sinB=c/sinC# and

cosine formulas #a^2=b^2+c^2-2bccosA#, #b^2=c^2+a^2-2cacosB# and #c^2=a^2+b^2-2abcosC#.

Here we are given #a=22#, #b=18# and #m/_C=130^@#, hence

#c^2=22^2+18^2-2xx22xx18xxcos130^@#

#=484+324-792xx(-0.6428)=808+509.0976=1317.0976#

i.e. #c=36.29# and using sine formula, we get

#22/sinA=18/sinB=36.29/(sin130^@)=36.29/0.766=47.376#

Hence, #sinA=22/47.376=0.4644# and #A=27.67^@#

and #sinB=18/47.376=0.3799# and #B=22.33^@#

Solution is #a=22#, #b=18#, #c=36.29#,

#m/_A=27.67^@#, #m/_B=22.33^@# and #m/_C=130^@#