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.

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