760. What are the sines, cosines, and tangents of the angles $30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$?
$\triangle$ In a right-angled triangle with an angle of $30^{\circ}$, the smaller leg is half the hypotenuse (completing to an equilateral triangle, problem 167), so $\sin 30^{\circ}=1 / 2$. The larger leg can be found using the Pythagorean theorem: if the hypotenuse is 1, then it is

$$
\cos 30^{\circ}=\sqrt{1^{2}-\left(\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}
$$

Their ratio, $\operatorname{tg} 30^{\circ}$, is therefore $1 / \sqrt{3}$.

For an angle of $60^{\circ}$, everything is the opposite: $\cos 60^{\circ}=1 / 2$, $\sin 60^{\circ}=\sqrt{3} / 2, \operatorname{tg} 60^{\circ}=\sqrt{3}$.

Finally, for a right-angled triangle with an angle of $45^{\circ}$ (an isosceles right-angled triangle, half of a square), we have

$$
\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}
$$

(the diagonal of a square is $\sqrt{2}$ times longer than its side), and $\operatorname{tg} 45^{\circ}=1 . \triangleleft$