Answer: [tex]\bold {1)\ \cos\ \theta}=\dfrac{\sqrt2}{2}\qquad 2)\ \tan \theta =\dfrac{1}{3}\qquad 3)\ \cos\ \theta=\dfrac{3\sqrt{13}}{13}\qquad 4)\ \cos\ \theta = \dfrac{2\sqrt5}{5}}[/tex]
Step-by-step explanation:
Pythagorean Theorem is: a² + b² = c², where "c" is the hypotenuse
[tex]1)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt2}\quad =\large\boxed{\dfrac{\sqrt2}{2}}[/tex]
Note: 2² + 2² = hypotenuse² → hypotenuse = 2√2
[tex]2)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{2}{2\sqrt2}\quad =\dfrac{5}{15}\quad \rightarrow \large\boxed{\dfrac{1}{3}}[/tex]
Note: hypotenuse not needed for tan θ
[tex]3)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3}{\sqrt{13}}\quad =\large\boxed{\dfrac{3\sqrt{13}}{13}}[/tex]
Note: 2² + 3² = hypotenuse² → hypotenuse = √13
[tex]4)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{2\sqrt5}\quad =\large\boxed{\dfrac{2\sqrt5}{5}}[/tex]
Note: hypotenuse given in problem
