# Problems based on triangles and trigonometry.

In an acute triangle $$x,y,z$$ are the given angles where $$\cos x=\tan y$$, $$\cos y = \tan z$$ and $$\cos z = \tan x$$. Find the sum of sines in the triangle.

Could be done by substituting values of $$\sin$$ function but in vain. Can anyone please help me?

• Normally you should show your efforts so far. May 19, 2020 at 19:29
• The sum of sines in the triangle does not have any sense, if such triangle does not exist. May 20, 2020 at 3:47

Hint:

Let $$a=\tan^2 x, b =\tan^2y, c=\tan^2 z$$. By squaring and using $$\cos^2x =\frac{1}{1+\tan^2x}$$, we have $$b=\frac{1}{1+a} \\ c=\frac{1}{1+b} \\ a=\frac{1}{1+c}$$ By substituting $$c$$ and $$b$$ you get a quadratic in $$a$$ : $$a^2+a-1=0 \implies a=\frac{\sqrt 5-1}{2}$$ and from here you get $$\sin x =\sqrt{\frac{\sqrt 5-1}{\sqrt 5+1}}$$ by some simple trig identities. You can find $$\sin y$$ and $$\sin z$$ in a similar way.

• The hint is nice, but the solution is (wrong) incomplete. The answer implies that $\tan x=\tan y=\tan z=\sqrt{\tfrac12\,(\sqrt 5-1)}$. What kind of a triangle is that? May 20, 2020 at 3:39
• @g.kov As the question, so the answer. May 20, 2020 at 8:29

The question states that it's given that in an acute triangle $$x,y,z$$ are the given angles where

\begin{align} \cos x&=\tan y \tag{1}\label{1} ,\\ \cos y&=\tan z \tag{2}\label{2} ,\\ \cos z&=\tan x \tag{3}\label{3} . \end{align}

There are many ways to prove that there is no valid triangle with such properties.

For one is the incomplete answer which implies that it follows from \eqref{1}-\eqref{3} that \begin{align} \tan x&=\tan y=\tan z= \cos x=\cos y=\cos z= =\sqrt{\tfrac12\,(\sqrt 5-1)} , \end{align}

which is absurd.

Another way is: rewriting \eqref{1}-\eqref{3} as

\begin{align} \cos x\cos y&=\sin y \tag{4}\label{4} ,\\ \cos y\cos z&=\sin z \tag{5}\label{5} ,\\ \cos z\cos x&=\sin x \tag{6}\label{6} , \end{align}
so \begin{align} \sin x+\sin y+\sin z &= \cos x\cos y+\cos y\cos z+\cos z\cos x \tag{7}\label{7} . \end{align}

Using known identities

\begin{align} \sin x+\sin y+\sin z&=u \tag{8}\label{8} \end{align}

and

\begin{align} \cos x\cos y+\cos y\cos z+\cos z\cos x &=\frac{u^2+v^2}4-1 \tag{9}\label{9} , \end{align}

where $$u=\rho/R$$, $$v=r/R$$ and $$\rho,r,R$$ are the semiperimeter, inradius and circumradius of given triangle (if such exists).

From equations \eqref{7}-\eqref{9},

\begin{align} \frac{u^2+v^2}4-1=v ,\\ u&=2+\sqrt{8-v^2} , \end{align}

and this expression for

\begin{align} u>\tfrac{3\sqrt3}2 &=\max_{v\in[0,1/2]}u(v) , \end{align}

that is, there is no a pair $$(u,v)$$ that simultaneously agree with \eqref{7}-\eqref{9} and represent a valid triangle.