MHB Km8,38 Find the value of \theta if it exists

[karush]

Find the value of $\theta$ if it exists
$$\theta=\tan^{-1}\sqrt{3}$$
rewrite
$$\tan(\theta)=\sqrt{3}$$
using $\displaystyle\tan\theta = \frac{\sin\theta}{\cos\theta}$ then if $\displaystyle\theta = \frac{\pi}{3}$
$$\displaystyle\frac{\sin\theta}{\cos\theta}
=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}
=\sqrt{3}$$ok I think this is a little awkward since it is observing the unit circle to see what will work
so was wondering if there is a more proper way.

 

[HallsofIvy]

My first thought would be to use a calculator (or, if you are as old as I am, a table of trig functions) but I suspect that is not what you mean. For $tan(\theta)= \sqrt{3}$, imagine an equilateral triangle with all three sides of length 2. Draw a perpendicular from one vertex to the opposite side. That will also bisect the opposites side and bisect the vertex angle. So it divides the equilateral triangle into two right triangles, each with hypotenuse of length 2 and one leg of length 1. By the Pythagorean theorem, The third sides, the altitude of the equilateral triangle, has length $\sqrt{2^2- 1^2}= \sqrt{3}$. So the tangent of the angle opposite that side of length $\sqrt{3}$ is $\frac{\sqrt{3}}{1}= \sqrt{3}$. Of course that angle is one of the original angles of the equilateral triangle so 60 degrees or $\pi/3$ radians.

 

[karush]

I'm 74

 

[HallsofIvy]

Just a young guy then! You'd be surprised what geezers most of us are. Nothing else to do, I guess.

 

[karush]

we have to let the young (under 70) know our brain didn't implode

just can't remember a D*** thing anymore...