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.


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