Answer:
sin angle B= square root of 5 / 5
Step-by-step explanation:
took the test and got it right!
The correct option is B. sin m∠B = square root of 5 over 5 .
Given [tex]\Delta ABC[/tex] is a right triangle.
The measure of Ac is 3, the measures of AB is 6 and the measure of BC is [tex]3\sqrt{5}[/tex].
Now, it is clear here
[tex](3\sqrt{5}) ^{2} = (3)^2+(6)^2\\[/tex]
[tex]9\times5=9+36\\45=45\\[/tex]
Hence the side BC will be the hypotenuse of the right triangle ABC and we know that the angle opposite to hypotenuse will be the right angle of the triangle.
Hence angle A is the right angle here.
We know that, from trigonometric ratio,
[tex]COS\ B=\dfrac{Base}{Hypotenuse}[/tex]
[tex]COS\ B=\dfrac{AB}{BC}[/tex]
[tex]COS \ B=\dfrac{6}{3\sqrt{5} }[/tex]
[tex]COS B=\dfrac{2}{\sqrt{5} } \\[/tex]
Or we can write,
[tex]COS B= \dfrac{2\sqrt{5} }{5}[/tex]
Similarly,
[tex]SIN \ B= \dfrac{AC}{BC} \\\\SIN \ B=\dfrac{ 3}{3\sqrt{5} }[/tex]
Or,[tex]SIN B= \dfrac{1}{\sqrt{5} }[/tex]
[tex]SIN B= \dfrac{\sqrt{5} }{5}[/tex]
Hence The correct option regarding angle B is option B. sin m∠B = square root of 5 over 5 .
For more details follow the link:
https://brainly.com/question/2263981
