Question

Os lados de um triagulo medem a=2√3,b=2√2, c=√6+√2 .Determine asbmedidas dos angulos internos

Answer 1

Lei dos Cossenos
a² = b² + c² - 2bc. cos â
(2√3)² = (2√2)² + (√6 + √2)² - 2.(2√2).(√6 + √2). cos â
12 = 8 + 6 + 2√12 + 2 - (4√2).(√6 + √2). cos â
12 = 16 + 2√4√3 - (4√12 +8). cos â
12 - 16 = 4√3 - ( 4√3√4 + 8) . cos â
- 4 - 4√3 = - ( 8√3 + 8) . cos â
-4 ( 1 + √3) = - 8 ( 1 + √3) . cos â
-4/-8 = cos â 
cos â = 1/2 , logo â = 60°

Pela lei dos senos
a/sen â = b/ sen ^b       sen â = sen 60 ° = √3/2
2√3/√3/2 = 2√2/sen b
2√3 sen b = √3/2. 2√2
sen b = √2/2
b = 45°

 a + b + c = 180º
60 + 45 + c = 180
c + 105 = 180
c = 75