Answer:
[tex]\cos(A)=\dfrac{3\sqrt{10}}{10}[/tex]
[tex]\sin(A)=\dfrac{\sqrt{10}}{10}[/tex]
[tex]\tan(A)=\dfrac13[/tex]
Step-by-step explanation:
Calculate the hypotenuse of ΔABC using Pythagoras' Theorem: [tex]a^2+b^2=c^2[/tex] (where a and b are the legs and c is the hypotenuse of a right triangle)
Given:
[tex]\implies 2^2+6^2=c^2\\\\\implies 4 + 36 = c^2\\\\\implies c^2=40\\\\\implies c=\sqrt{40} \\\\\implies c=2\sqrt{10}[/tex]
Trig ratios:
[tex]\sin(\theta)=\dfrac{O}{H} \ \ \ \ \cos(\theta)=\dfrac{A}{H} \ \ \ \ \tan(\theta)=\dfrac{O}{A}[/tex]
where [tex]\theta[/tex] is the angle, O is the side opposite the angle, A is the side adjacent the angle, H is the hypotenuse in a right triangle.
[tex]\cos(A)=\dfrac{6}{2\sqrt{10} }=\dfrac{3\sqrt{10}}{10}[/tex]
[tex]\sin(A)=\dfrac{2}{2\sqrt{10} }=\dfrac{\sqrt{10}}{10}[/tex]
[tex]\tan(A)=\dfrac{2}{6}=\dfrac13[/tex]
