# GEOM 2 | Lesson 3 | Practice (Ratios for Special Right Triangles) Solutions

1.  $$\sin 30˚ = \dfrac{1}{2}$$ ; $$\sin 30˚ = \dfrac{\dfrac{1}{\sqrt 3}}{\dfrac{2}{\sqrt 3}}$$ ;

$$\sin 30˚ = \dfrac{1}{2}$$

Note that the ratios are the same:      $$\dfrac{1}{2}$$

$$\cos 30˚ = \dfrac{\sqrt 3}{2}$$;  $$\cos 30˚ = \dfrac{1}{\dfrac{2}{\sqrt 3}}$$;

$$\cos 30˚ = \dfrac{\dfrac{\sqrt 3}{2}}{1}$$

Note that the ratios are the same:  $$\dfrac{\sqrt 3}{2}$$

2.  $$\sin 60˚ = \dfrac{\sqrt 3}{2}$$ ;$$\sin 60˚= \dfrac{1}{\dfrac{2}{\sqrt 3}}$$;  $$\sin 60˚ = \dfrac{\sqrt 3}{2}$$.  Note:  The ratios are all $$\dfrac{\sqrt 3}{2}$$.

$$\cos 60˚ = \dfrac{1}{2}$$; $$\cos 60˚ = \dfrac{\dfrac{1}{\sqrt 3}}{\dfrac{2}{\sqrt 3}}$$;

$$\cos 60˚ = \dfrac{1}{2}$$.

Note that the ratios are the same:      $$\dfrac{1}{2}$$

3.  $$\sin 45˚ = \dfrac{1}{\sqrt 2}$$; $$\cos 45˚ = \dfrac{1}{\sqrt 2}$$  or $$\dfrac{\sqrt 2}{2}$$

$$\sin 45˚ =\dfrac{1}{\sqrt 2}$$; $$\cos 45˚=\dfrac{1}{\sqrt2}$$ or $$\dfrac{\sqrt 2}{2}$$

4.  The legs of the right triangle are the same length.

5.

Angle sin cos tan csc sec cot
Θ
30° $$\dfrac{1}{2}$$ $$\dfrac{\sqrt3}{2}$$ $$\dfrac{\sqrt3}{3}$$ 2 $$\dfrac{2}{√3}$$ $$\sqrt3$$
60° $$\dfrac{\sqrt3}{2}$$ $$\dfrac{1}{2}$$ $$\sqrt3$$ $$\dfrac{2\sqrt3}{3}$$ 2 $$\dfrac{\sqrt3}{3}$$
45° $$\dfrac{\sqrt2}{2}$$ $$\dfrac{\sqrt2}{2}$$ 1 $$\sqrt2$$ $$\sqrt2$$ 1

6.    Note that $$\sin 30˚ = \cos 60˚$$; $$\sin 60˚ = \cos 30˚$$; $$\tan 30˚ = \dfrac1{\tan 60˚}$$;

$$\sin 45˚ = \cos45˚$$ ;

Reciprocals:  $$\csc Θ = \dfrac{1}{\sin Θ}$$; $$\sec Θ =\dfrac{1}{\cos Θ}$$;

$$\cot Θ = \dfrac{1}{\tan Θ}$$

7.  You probably learned SOHCAHTOA in high school.