Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$

Question

What is $$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$ for a triangle with sides $2$, $3$, and $4$?

One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc)$ to get the cosines. But that's lots of calculate.

Is there a better way to get the answer? Thanks!

Answer 1

There are well-known identities for $\triangle ABC$ with the angles $A,B,C$, sides $a,b,c$, semiperimeter $\rho=\tfrac12(a+b+c)$, area $S$, radius $r$ of inscribed and radius $R$ of circumscribed circles,

\begin{align} \sin A+\sin B+\sin C &=\frac\rho{R} \tag{1}\label{1} ,\\ \cos A+\cos B+\cos C &=\frac{r+R}{R} \tag{2}\label{2} , \end{align}

so

\begin{align} x&= \frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C} =\frac{\rho}{r+R} \tag{3}\label{3} , \end{align}

we also know that

\begin{align} R&=\frac{abc}{4S} ,\\ r&=\frac{S}{\rho} ,\\ S&=\tfrac14\sqrt{4(ab)^2-(a^2+b^2-c^2)^2} ,\\ \end{align}

thus we can find that for $a=2,b=3,c=4$ \begin{align} \rho&=\frac{9}{2} ,\\ S&=\frac{3\sqrt{15}}{4} ,\\ R&=\frac{8\sqrt{15}}{15} ,\\ r&=\frac{\sqrt{15}}{6} ,\\ x&=\frac{\rho}{r+R} =\frac{3\sqrt{15}}{7} \approx 1.6598500 . \end{align}

Answer 2

Using an application of the Inscribed Angle Theorem, we get $$ \begin{align} 2R\sin(A)=a\tag{1a}\\ 2R\sin(B)=b\tag{1b}\\ 2R\sin(C)=c\tag{1c} \end{align} $$ where $R$ is the radius of the circumcircle.

Furthermore, with $s=\frac{a+b+c}2$, $$ \begin{align} \text{Area} &=\sqrt{s(s-a)(s-b)(s-c)}\tag2\\[3pt] &=\frac12bc\sin(A)\tag3\\ &=\frac{abc}{4R}\tag4 \end{align} $$ Explanation:
$(2)$: Heron's Formula
$(3)$: triangular area given by Cross Product
$(4)$: apply $\text{(1a)}$ to $(3)$

Therefore, $$ \begin{align} \sin(A)+\sin(B)+\sin(C) &=\frac{a+b+c}{2R}\tag5\\ &=\frac{4s\sqrt{s(s-a)(s-b)(s-c)}}{abc}\tag6 \end{align} $$ Explanation:
$(5)$: apply $\text{(1a)}$, $\text{(1b)}$, and $\text{(1c)}$
$(6)$: get $R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$ from $(2)$ and $(4)$

The Law of Cosines says $$ \begin{align} \cos(A)&=\frac{b^2a+c^2a-a^3}{2abc}\tag{7a}\\ \cos(C)&=\frac{c^2b+a^2b-b^3}{2abc}\tag{7b}\\ \cos(C)&=\frac{a^2c+b^2c-c^3}{2abc}\tag{7c} \end{align} $$ Adding these and factoring yields $$ \begin{align} \cos(A)+\cos(B)+\cos(C) &=\frac{(a+b-c)(a-b+c)(-a+b+c)}{2abc}+1\tag8\\ &=\frac{4(s-a)(s-b)(s-c)}{abc}+1\tag9 \end{align} $$ Combining $(6)$ and $(9)$ gives $$ \frac{\sin(A)+\sin(B)+\sin(C)}{\cos(A)+\cos(B)+\cos(C)}=\frac{4s\sqrt{s(s-a)(s-b)(s-c)}}{4(s-a)(s-b)(s-c)+abc}\tag{10} $$

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Plugging $(a,b,c)=(2,3,4)$ into $(10)$ gives $$ \begin{align} \frac{\sin(A)+\sin(B)+\sin(C)}{\cos(A)+\cos(B)+\cos(C)} &=\frac{4\cdot\frac92\sqrt{\frac92\cdot\frac52\cdot\frac32\cdot\frac12}}{4\cdot\frac52\cdot\frac32\cdot\frac12+2\cdot3\cdot4}\\ &=\frac{3\sqrt{15}}{7}\tag{11} \end{align} $$