# How to determine angle without using trigonometry? [closed]

If we know that $\sin \theta = \frac{1}{\sqrt2}$ and $\cos \theta = \frac{1}{\sqrt2}$ then how can we determine that the right triangle with sides $1$, $1$ and $\sqrt{2}$ also has angles of $45º$ just by using geometry and some algebra, without using trigonometric identities (but only knowing the relationships of the sides given by the sine and cosine)?

Update: I retrace my question. My original intent was to find the angles without using trigonometric identities, but just based on the relationships of the sides given by the sine and cosine. I think such would be impossible, however. But using the laws of cosines or sines is, of course, a very well-known approach.

You have $\sin \theta =\cos \theta = \sin (\frac \pi 2 - \theta)$. As $\sin \theta$ is increasing in this range, this gives $\theta=\frac \pi 2-\theta, \theta=\frac \pi 4$

You can use the fact that base angles of an isosceles triangle are equal. You know the $1-1-\sqrt 2$ triangle has a right angle by Pythagoras, so the other two angles have to be $45^\circ$

• @sequence: your first line uses the sine and cosine functions. Commented Aug 12, 2018 at 22:57
• I retrace my question. My original intent was to find the angles without using trigonometric identities, but just based on the relationships of the sides given by the sine and cosine. I think such would be impossible, however. But using the laws of cosines or sines is, of course, a very well-known approach. Commented Aug 13, 2018 at 0:19

From the diagram, we have $a=1,b=\sqrt{2},c=1$

We already know $\angle B=90^{\circ}$

To find the angle $C$ we can use law of cosines, $$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$$$\cos C=\dfrac{1+2-1}{2\sqrt{2}}$$ $$\cos C=\dfrac{1}{\sqrt{2}}$$ $$\angle C=45^{\circ}$$ Similarly, we get $\angle B=45^{\circ}$

• But what if we didn't know that $\cos C = \frac{1}{\sqrt2}$ implies $C=45º$? Commented Aug 13, 2018 at 0:25
• @sequence The other two angles are equal by self-similarity of the triangle. The angles add up to $180^\circ$. Therefore $C$ and $A$ are both $45^\circ$. Commented Aug 13, 2018 at 5:20