Question

If in a triangle ABC A=2B then a^(2)-b^(2)= a) ab b) bc c) ca d) (ab)/(2)​

Answer 1

Given :  in a triangle ABC ∠A=2∠B

To Find : a² - b²

Solution:

∠A  = 2∠B

Taking sin both sides

=> Sin ∠A = sin 2∠B

=> Sin ∠A = 2 sin  ∠B . Cos ∠B

a/SinA  = b/SinB = c/SinC = k

=> SinA =a/k

   SinB = b/k  

b² = a² + c² - 2acCos∠B

=> Cos∠B = ( a² +c² - b²) / 2ac

=> a/k    = 2 (b/k) ( a² +c² - b²) / 2ac

=> a²c = a²b + b(c² - b² )

=> a²c - a²b - b(c² - b² ) = 0

=> a² ( c - b)  - b(c - b)(c + b) = 0

=>  (c - b) (a²   - bc - b² ) = 0

=> c = b or  a²   - b² = bc

a²   - b² = bc

a²   - b² = bc   is the correct option

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