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# 17thKVS-JMO2014-Solutions4

## 17thKVS-JMO2014-Solutions4

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4Q) In a triangle ABC, angle A is twice the angle B. Show that $$a^{2}$$ = b(b+c) , where a, b and c are the sides opposite to angle A, B and C respectively.

Solution :

Given    A = 2B

Using cosine rule :

a² = b² + c² - 2bc cos(A)  , we get

a² = b² + c² - 2bc cos(2B)------------------(1)

b² = c² + a² - 2ac cos(B) -----------------(2)

(1) - (2) we get

a² - b² = b² - a² +2c (acos(B) - b cos(2B))

2a² - 2b² = 2c ( a cos(B) - b cos(2B))

a² - b² = c(a cos(B) - b(2cos²(B)-1))

$\left [ \because \cos \left ( 2B \right ) = 2\cos ^{2}\left ( B \right )-1\right ]$

a² - b² = c(acos(B)- 2bcos²(B) +b)

a² - b² = c ( 2R sin(A) cos(B) - 2(2R sin(B))cos²(B) +b)

$\left ( \because a=2R\sin (A), b=2R\sin (B) \right )$

a² - b² = c ( 2Rsin(A) cos(B) - 4Rsin(B) cos²(B) + b)

a² -b² = bc + 2c R cos(B)(sin(A) - 2sin(B)cos(B))

a² - b² = bc + 2c R cos(B) (sin(2B)-sin(2B))

a² - b² = bc + 2c R cos(B)(0)

a² -b² = bc + 0

a² = b² + bc

a² = b(b+c)