# 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:
$\displaystyle \frac{{\sin a}}{{\sin b}} = 2\cos c$
• 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:
$\displaystyle \frac{{\sin a}}{{\sin b}} = 2\cos c$

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

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

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

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

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

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

$\displaystyle \Rightarrow \tan C = \tan B$

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

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

$\displaystyle \Rightarrow CP = \tfrac12a$

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

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