A cute problem in Trigonometry or pure plane geometry

Mathematics Hothouse

**Problem.**

In a triangle ABC, $latex angle A$ is twice $latex angle B$. Show that $latex a^{2}=b(b+c)$. (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.

$latex A=2B Longleftrightarrow A-B=B Longleftrightarrow sin{(A-B)}=sin{B} Longleftrightarrow sin{(A-B)}sin{(A+B)}=sin{B}sin{C} Longleftrightarrow sin^{2}{A}-sin^{2}{B}=sin{B}sin{C} Longleftrightarrow (2Rsin{A})^{2}-(2Rsin{B})^{2}=(2Rsin{B})(2Rsin{C}) Longleftrightarrow a^{2}-b^{2}=bc Longleftrightarrow a^{2}=b(b+c)$

View original post

### Like this:

Like Loading...

*Related*