In △ABC , m∠A=53°,m∠B=17°, and a=27. Find the perimeter of the triangle.
1 answer:
Using the Law of Sines (sinA/a=sinB/b=sinC/c) and the fact that all triangles have a sum of 180° for their angles.
The third angle is C is 180-53-17=110°
27/sin53=b/sin17=c/sin110
b=27sin17/sin53, c=27sin110/sin53
And the perimeter is a+b+c so
p=27+27sin17/sin53+27sin110/sin53 units
p≈68.65 units (to nearest hundredth of a unit)
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