OSTP Grade 11 CCRA Trigonometry. Practice it free.

$\dfrac{\pi}{6} \times \dfrac{180°}{\pi} = 30°$.

The fundamental Pythagorean identity is $\sin^{2}\theta + \cos^{2}\theta = 1\textsf{.}$

The reference angle for $\dfrac{7\pi}{4}$ is $\dfrac{\pi}{4}$. Tangent is negative in Quadrant IV, so $\tan\left(\dfrac{7\pi}{4}\right) = -\tan\left(\dfrac{\pi}{4}\right) = -1$.

The domain of $\arcsin(x)$ is $[-1,\, 1]$.

The short leg is half the hypotenuse: $\dfrac{16}{2} = 8$.

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\dfrac{7}{25}}{\dfrac{24}{25}} = \dfrac{7}{25} \cdot \dfrac{25}{24} = \dfrac{7}{24}$.

In a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ triangle, the short leg is half the hypotenuse: $\dfrac{16}{2} = 8\textsf{.}$

Both legs of a $45$-$45$-$90$ triangle are equal, so the other leg is $7$.