# GEOM 2| Lesson 3| Practice Solutions

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

$$\sin 30˚ = \frac{1}{2}$$

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

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

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

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

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

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

$$\cos 60˚ = \frac{1}{2}$$.

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

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

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

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

5.

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

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

$$\sin 45˚ = \cos45˚$$ ;

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

$$\cot Θ = \frac{1}{\tan Θ}$$

7.  You probably learned SOHCAHTOA in high school.