In a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ is larger than $s_2$

Question

While solving a totally different problem , I made observation that in a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ say to $θ_1$ is larger than angle opposite to $s_2$ say $θ_2$.

I verified by drawing several triangles . Can someone tell if it's right and prove it ?

Answer 1

It is a well-known property. It is in fact Euclid's Proposition 18 in his first book of the Elements. See this link for Euclid's argument.

Answer 2

Following formula shows that your claim is correct:

$\frac a{sin (\alpha)}=\frac b{sin (\beta) }=\frac c{sin (\gamma)}=2R$

Where R is radius of circumcircle of triangle ABC with sides a, b and c and opposite angle to these sides $\alpha, \beta, \gamma$.

Answer 3

In the diagram given above, $AB>BC$. Now, if this is true, then there must be a point $D$ between points $A$ and $B$ on segment $AB$ such that $AD=AC$. Draw $DC$.

Now, in $\triangle ADC$, since $AD=DC$, $\angle ADC=\angle ACD$[opposite angles of equal sides are equal].

$\angle ACD<\angle ACB$, since point $D$ is between $A$ and $B$.

Also, $\angle ADC=\angle DBC+\angle BCD>\angle DBC=\angle ABC$.

Thus, $\angle ACB>\angle ACD=\angle ADC>\angle ABC$.

$\angle ACB$ and $\angle ABC$ are the opposite angles of side $AB$ and $AC$ respectively. Hence, we have proved that $\angle ACB=$ opposite angle of longer side $AB>\angle ABC=$ opposite angle of shorter side $AC$.