Question: **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 $x$?

Answer: 
$\theta = \tan^{-1}(2), \beta = \tan^{-1}(3)\implies \tan(\theta) = 2, \tan(\beta) = 3\implies \tan(\beta+\theta)=\dfrac{2+3}{1-2\cdot 3}= -1\implies \beta+\theta= 3\pi/4\implies \gamma = \pi/4$
Three angles of a triangle sum to 180 degrees

Consider this figure. (the grid is a square grid.)

[![enter image description here](https://i.stack.imgur.com/8REGU.png)](https://i.stack.imgur.com/8REGU.png)

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.