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In a triangle ABC,angleA is twice the angleB. show that a 2 =b[b+c],where a,b and c are the sides opposite to angle A,B and C.

In a triangle ABC,angleA is twice the angleB. show that a2  =b[b+c],where a,b and c are the sides opposite to angle A,B and C.

Grade:10

1 Answers

Satyajit Samal
askIITians Faculty 34 Points
8 years ago
A=2B \Rightarrow A-B=B \Rightarrow \sin \left ( A-B \right )= \sin B
Multiplying both sides by\sin C, we get
\sin C \sin \left ( A-B \right )= \sin B \sin C \\ \Rightarrow \sin \left ( \pi -\left ( A+B \right ) \right ) \sin \left ( A-B \right )= \sin B \sin C \\ \Rightarrow \sin \left ( A+B \right ) \sin \left ( A-B \right )= \sin B \sin C \\ \Rightarrow \sin^{2}A - \sin^{2}B = \sin B \sin C \\ \Rightarrow a^2 - b^2 = bc \left ( \because \sin A = Ka,\; \sin B = Kb,\; \sin C = Kc \right )\\ \RIghtarrow a^2 = b\left ( b+c \right )
Hence proved.

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