Appendix B - Trigonometry Review - Exercise Set B - Page A23: 5

$${\text{See explanation}}$$

$$\eqalign{ & 5a) \cr & {\text{From the triangle we can see that:}} \cr & {\text{Opposite side to }}\theta = \sqrt {{5^2} - {2^2}} = \sqrt {21} \cr & {\text{Adjacent side to }}\theta = 2 \cr & {\text{Hypotenuse }}\theta = 5 \cr & {\text{Thus}}{\text{,}} \cr & \cr & {\text{sin}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{{\sqrt {21} }}{5},\,\,\,\,\,{\text{csc}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{5}{{\sqrt {21} }}\,\, \cr & {\text{cos}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{2}{5},\,\,\,\,\,{\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{5}{2}\,\, \cr & {\text{tan}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{{\sqrt {21} }}{2},\,\,\,\,\,{\text{cot}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{2}{{\sqrt {21} }}\,\, \cr & \cr & 5b) \cr & {\text{From the triangle we can see that:}} \cr & {\text{Opposite side to }}\theta = 3 \cr & {\text{Adjacent side to }}\theta = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & {\text{Hypotenuse }}\theta = 4 \cr & {\text{Thus}}{\text{,}} \cr & \cr & {\text{sin}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{3}{4},\,\,\,\,\,{\text{csc}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{4}{3}\,\, \cr & {\text{cos}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{{\sqrt 7 }}{4},\,\,\,\,\,{\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{4}{{\sqrt 7 }}\,\, \cr & {\text{tan}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{3}{{\sqrt 7 }},\,\,\,\,\,{\text{cot}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{{\sqrt 7 }}{3} \cr & \cr & 5c) \cr & {\text{From the triangle we can see that:}} \cr & {\text{Opposite side to }}\theta = 3 \cr & {\text{Adjacent side to }}\theta = \sqrt {{4^2} - {3^2}} = 1 \cr & {\text{Hypotenuse }}\theta = \sqrt {{1^2} + {3^2}} = \sqrt {10} \cr & {\text{Thus}}{\text{,}} \cr & \cr & {\text{sin}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{3}{{\sqrt {10} }},\,\,\,\,\,{\text{csc}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{{\sqrt {10} }}{3}\,\, \cr & {\text{cos}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{1}{{\sqrt {10} }},\,\,\,\,\,{\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{{\sqrt {10} }}{1}\,\, \cr & {\text{tan}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = 3,\,\,\,\,\,{\text{cot}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{1}{3} \cr} $$