Answer:
[tex]cos B = \dfrac{2\times \sqrt5}{5}[/tex] is the correct answer.
Step-by-step explanation:
We are given a right angled [tex]\triangle ABC[/tex] with the side measurements as:
AB = 6
AC = 3
BC = 3[tex]\sqrt5[/tex]
In a right angled triangle, the angle of [tex]90^\circ[/tex] is the largest angle and the side opposite to the right angle is the largest.
We are given that the side BC = 3[tex]\sqrt5[/tex] which has a value greater than 6 which means side BC is the largest side.
And angle opposite to BC is [tex]\angle A[/tex] is the largest i.e.
[tex]\angle A=90^\circ[/tex]
Please refer to the attached image for the given dimensions of the triangle.
Now, we can apply trigonometric rules to easily find out the value of [tex]\angle B[/tex]
[tex]cos \theta = \dfrac{Base}{Hypotenuse}\\OR\\cos B = \dfrac{AB}{BC}\\\Rightarrow cos B = \dfrac{6}{3\sqrt5}\\\Rightarrow cos B = \dfrac{2}{\sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{\sqrt5 \times \sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{5}[/tex]
