In a triangle $\Delta ABC$, find angle A if $r:R:r_1=2:5:12$

Question

Context: $r_1$ is the ex circle opposite angle A, $r$ is the incircle and $R$ is the circumcircle.

Let the proportionality constant be $x$

$$4R\sin A/2 \sin B/2 \sin C/2 =2x$$ $$R=5x$$ $$4R\sin A/2 \cos B/2 \cos C/2 = 12x$$

Dividing 1 and 3 $$\tan B/2 \tan C/2 =\frac 16$$

How should I solve further.

Answer 1

Your way to go works fine, you need after the attempt from the OP $$ \begin{aligned} \sin\frac A2\sin \frac B2\sin \frac C2 &=\frac {2x}{4\cdot 5x}=\frac 1{10}\ ,\\ \sin\frac A2\cos \frac B2\cos \frac C2 &=\frac {12x}{4\cdot 5x}=\frac 6{10}\ , \end{aligned} $$ just one more small step: $$ \frac 12= \frac 6{10}-\frac 1{10} =\sin\frac A2 \underbrace{ \left( \cos \frac B2\cos \frac C2 - \sin \frac B2\sin \frac C2 \right)} _{\displaystyle=\cos\frac{B+C}2=\sin\frac A2} =\sin^2\frac A2 \ . $$ So $\frac A2$ is $45^\circ$, the angle with the sine $\frac 1{\sqrt 2}$

Answer 2

The area of the triangle is

$$Area = \frac12r(a+b+c) = \frac12r_1(c+b-a)= \frac {abc}{4R}$$

which leads to

$${b+c}=\frac{r_1+r}{r_1-r}a= \frac75a,\>\>\>\>\>bc=\frac{48}{25}R^2=\frac{12}{25}\frac{a^2}{\sin^2A}\tag 1$$

Reexpress the cosine rule for the triangle $a^2=b^2+c^2-2bc\cos A $ as

$$(b+c)^2-a^2=2bc(1+\cos A )$$

Substitute (1) to get

$$\cos A(1+\cos A)=0$$

which yields $\cos A=0$, i.e. $A=90^\circ$.

Answer 3

Let \begin{align} r&=2\,k ,\quad R=5\,k ,\quad r_a=12\,k \tag{1}\label{1} . \end{align}

We can find the radii $r_a,r_b$ of other two excircles, using known identities \begin{align} r_b+r_c &=r+4\,R-r_a \tag{2}\label{2} ,\\ \frac1{r_b}+\frac1{r_c} &=\frac1r-\frac1{r_a} \tag{3}\label{3} \end{align}

and assuming WLOG that $r_b\le r_c$,we get

\begin{align} 2+\frac{r_b}{r_c}+\frac{r_c}{r_b} &= \frac{25}6 ,\\ \frac{r_b}{r_c}&=\frac23 \tag{4}\label{4} . \end{align}

\eqref{4} combined with \eqref{3} gives

\begin{align} r_b&=4\,k ,\quad r_c=6\,k . \end{align}

Noe we can apply a neat expression fo find the $\cos\alpha$:

\begin{align} \cos\alpha&= \frac{r_br_c-rr_a}{r_br_c+rr_a} \end{align} and get the answer: $\cos\alpha=0$.

Similarly, the other two are

\begin{align} \cos\beta &= \frac{r_ar_c-rr_b}{r_ar_c+rr_b} =\frac45 ,\\ \cos\gamma &= \frac{r_ar_b-rr_c}{r_ar_b+rr_c} =\frac35 , \end{align}

and we can conclude that the triangle is similar to the famous $3-4-5$ right-angled triangle.