A problem related to solution of triangles

Problem.

In a triangle ABC, $\angle A$ is twice $\angle B$. Show that $a^{2}=b(b+c)$. (In fact, the converse is also true. Prove it!).

Proof.

Method I.

You can use plane geometry also. This is left to you as an exercise.

Method II.

You can use trigonometry also. We may use the sine rule for a triangle to dispose of both the implications simultaneously. $A=2B \Longleftrightarrow A-B=B \Longleftrightarrow \sin{(A-B)}=\sin{B} \Longleftrightarrow \sin{(A-B)}\sin{(A+B)}=\sin{B}\sin{C} \Longleftrightarrow \sin^{2}{A}-\sin^{2}{B}=\sin{B}\sin{C} \Longleftrightarrow (2R\sin{A})^{2}-(2R\sin{B})^{2}=(2R\sin{B})(2R\sin{C}) \Longleftrightarrow a^{2}-b^{2}=bc \Longleftrightarrow a^{2}=b(b+c)$

One Comment

1. nkpithwa
Posted July 23, 2015 at 10:56 am | Permalink | Reply

Reblogged this on Mathematics Hothouse and commented:

A cute problem in Trigonometry or pure plane geometry

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