# Solve the oblique triangle  side a=20,  side b=30, angle C=60degrees. Find the missing sides and angles. I know it's a SSA triangle, but I need step by step instructions on how to solve. You may find the length of misssing side c using the law of cosines such that:

`c^2 = a^2+b^2 - 2ab*cos hatC`

`c^2 = 20^2 + 30^2 - 2*20*30*cos 60^o`

`c^2 = 400 + 900 - 600 => c^2 = 700 => c = 10sqrt7`

You may find the measure of missing angle using the law of sines such that:

`c/sin hatC = b/sin hat B => bsin hatC = c sin hat B`

`30*sin60^o = 10sqrt7*sin hat B => 3sqrt3/2 = sqrt7*sin hat B`

`sin hat B= 3sqrt2/2sqrt7 => sin hat B = 3sqrt14/14 => hat B = 53^o`

Since evaluating the sum of interior angles of triangle yields `180^o` , you may find the missing angle hat A such that:

`hat A = 180^o - (60^o + 53^o) => hat A = 67^o`

Hence, evaluating the length and measures of missing side and angles, using the laws of cosines and sines, yields  `c = 10sqrt7, hat B = 53^o, hat A = 67^o .`

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