Nature of the triangle

• Dec 21st 2009, 09:12 AM
dhiab
Nature of the triangle
Determine the nature of the triangle ABC, the measurements of his angles of which are respectively, a, b, c such as:
$\frac{{\sin a}}{{\sin b}} = 2\cos c
$
• Dec 22nd 2009, 12:54 AM
Hello dhiab
Quote:

Originally Posted by dhiab
Determine the nature of the triangle ABC, the measurements of his angles of which are respectively, a, b, c such as:
$\frac{{\sin a}}{{\sin b}} = 2\cos c
$

It is isosceles, with $\angle B = \angle C$.

This may be proved by trigonometry formulae:
$\frac{\sin A}{\sin B}=2\cos C$

$\Rightarrow \frac{\sin(B+C)}{\sin B}= 2\cos C$, since $A = 180^o -(B+C)$

$\Rightarrow \frac{\sin B \cos C + \cos B \sin C}{\sin B}=2\cos C$

$\Rightarrow \cos C + \cot B \sin C = 2\cos C$

$\Rightarrow \cot B\sin C = \cos C$

$\Rightarrow \tan C = \tan B$

$\Rightarrow C = B$, since $B$ and $C$ both lie between $0^o$ and $180^o$
or, using the Sine Rule, which gives:
$\frac{\sin A}{\sin B}=\frac{a}{b}=2\cos C$

$\Rightarrow \cos C = \frac{\frac12a}{b}$
and then considering the attached diagram:
$P$ is the foot of the perpendicular from $A$ to $BC$. Then
$\cos C = \frac{CP}{AC}= \frac{\frac12a}{b}$

$\Rightarrow CP = \tfrac12a$

$\Rightarrow P$ is the mid-point of $CB$

$\Rightarrow \triangle ABC$ is isosceles with $AC = AB$.
$\frac{\sin A}{\sin B}=2\cos C\Rightarrow\sin A=2\sin B\cos C\Rightarrow$
$\Rightarrow \sin A=\sin(B+C)+\sin(B-C)\Rightarrow$
$\Rightarrow\sin A=\sin A+\sin(B-C)\Rightarrow\sin(B-C)=0\Rightarrow B=C$