Question: If two angles of a triangle $ABC$ are $\arctan 2$ and $\arctan 3$, what is the third angle?
My attempt: Let the third angle of the triangle $ABC$ be $x$.
$\therefore$ $\arctan 2+\arctan 3+ x=\pi$
$\implies x=\pi-(\arctan 2+\arctan 3)$
Now, there is a formula: $\arctan a +\arctan b=\arctan(\frac{a+b}{1-ab})$ where $ab<1$.
Here $ab=2\times 3=6$ which is greater than $1$.
So how do i solve this equation to obtain the value of $x$?
In a triangle $ABC$ we have $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ Therefore $$2+3+\tan C=2\cdot3\cdot\tan C$$ So, $$\tan C=1$$ and $C=\pi/4$.
$\theta = \tan^{-1}(2), \beta = \tan^{-1}(3)\implies \tan(\theta) = 2, \tan(\beta) = 3\implies \tan(\beta+\theta)=\dfrac{2+3}{1-2\cdot 3}= -1\implies \beta+\theta= 3\pi/4\implies \gamma = \pi/4$
Three angles of a triangle sum to 180 degrees
Consider this figure. (the grid is a square grid.)
Angles $A+B+C$ sum to $180.$
$A$ is an angle with measure $\arctan 3.$
$B$ has measure $\arctan 2$
What is the measure of angle $C$?
It is an isosceles right triangle.
