Question

Given triangle ABC, which equation could be used to find the measure of ∠B? right triangle ABC with AB measuring 6, AC measuring 3, and BC measuring 3 square root of 5

A. cos m∠B = square root of 5 over 5
B. sin m∠B = square root of 5 over 5
C. cos m∠B = square root of 5 over 2
D. sin m∠B = 2 square root of 5 all over 5

Answer 1

A

Explanation:

We have a right triangle ABC, and we're given the side lengths:

- AB = 6

- AC = 3

- BC = 3√5

To find the measure of angle ∠B, we can use trigonometric functions. In a right triangle, the cosine of an angle is defined as the adjacent side divided by the hypotenuse, and the sine of an angle is defined as the opposite side divided by the hypotenuse.

In this case, angle ∠B is the angle between sides AB (adjacent to angle ∠B) and BC (the hypotenuse). So, the equation we should use is:

**cos(∠B) = adjacent / hypotenuse = AB / BC**

Substituting the given values:

**cos(∠B) = 6 / (3√5) = 2 / √5 = 2√5 / 5**

None of the given answer choices match this result exactly, but the closest one is:

A. **cos(∠B) = √5 /