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 cos m∠B = 2 square root of 5 all over 5 sin m∠B = square root 5 cos m∠B = square root of 5 over 2 sin m∠B = 2 square root of 5 all over 5

Answer 1

Answer:

$cos B = \dfrac{2\times \sqrt5}{5}$ is the correct answer.

Step-by-step explanation:

We are given a right angled $\triangle ABC$ with the side measurements as:

AB = 6

AC = 3

BC = 3$\sqrt5$

In a right angled triangle, the angle of $90^\circ$ is the largest angle and the side opposite to the right angle is the largest.

We are given that the side BC = 3$\sqrt5$ which has a value greater than 6 which means side BC is the largest side.

And angle opposite to BC is $\angle A$ is the largest i.e.

$\angle A=90^\circ$

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 $\angle B$

$cos \theta = \dfrac{Base}{Hypotenuse}\\OR\\cos B = \dfrac{AB}{BC}\\\Rightarrow cos B = \dfrac{6}{3\sqrt5}\\\Rightarrow cos B = \dfrac{2}{\sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{\sqrt5 \times \sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{5}$