# How to prove this part of a problem?

In triangle $ABC$, $\angle A=x$, $\angle B=2x$ and $AC/BC= 1/2000$

Can you prove that $x$ must be in between $0$ and $90$ degrees, and that $\sin (x)$ cannot be $0$.

I can kind of explain this, but mostly with words... how can I prove this more kind of "mathematically"?" (This is what I've been thinking: there are $3x + \angle y = 180$, that's why x can't be bigger $90$. It can't be less than $0$ because then it wouldn't exist) And as for $\sin x$ not being $0$, that would mean that the $\angle x$ is $0$, and that can't be...)

• do you mind to explain your idea in words...?
– user87543
Apr 9, 2014 at 5:55

Not only can we prove X is between 0 and 90 degrees, we can prove it is between 0 and 60 degrees.

Call our 3 angles X, 2X, and Y. We know that X + 2X + Y = 180. Therefore:

3X + Y = 180
3X = 180 - Y
X = (180 - Y) / 3
X = 60 - Y/3

Since Y is guaranteed to be some positive value, X is 60 minus that value, and is greater than 0 by virtue of being an angle in a triangle. This puts it in the range of 0 to 90.

Like you originally stated, the proof that sin(x) =/= 0 is simple: by this graph, sin(x) only equals 0 at x=0 + 180° * k, where k is an integer. This never falls in our possible values for x (0 < x < 60) so sin(x) is never 0.

The angles of a triangle are all greater than $0$ and less than $180$, and since the sine of any angle in that range is positive, $\sin(x) = 0$ cannot be satisfied. And because the three angles of a triangle sum to $180$, the sum of two of them $A + B = 3 \cdot x$ is less than $180$, so $A = x$ is less than $60$.