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
$

Hello dhiab

Quote:

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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
$

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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 $\displaystyle AC = AB$.

Grandad

Another approach:

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

$\displaystyle \Rightarrow \sin A=\sin(B+C)+\sin(B-C)\Rightarrow$

$\displaystyle \Rightarrow\sin A=\sin A+\sin(B-C)\Rightarrow\sin(B-C)=0\Rightarrow B=C$