Find cos(2* angle ABC) 100 POINTS

3/5

Step-by-step explanation:

We need to use the trig identity that cos(2A) = cos²A - sin²A, where A is an angle. In this case, A is ∠ABC. Essentially, we want to find cos∠ABC and sin∠ABC to solve this problem.

Cosine is adjacent ÷ hypotenuse. Here, the adjacent side of ∠ABC is side BC, which is 4 units. The hypotenuse is 2√5. So, cos∠ABC = 4/2√5 = 2/√5.

Sine is opposite ÷ hypotenuse. Here, the opposite side of ∠ABC is side AC, which is 2 units. The hypotenuse is still 2√5. So sin∠ABC = 2/2√5 = 1/√5.

Now, cos²∠ABC = (cos∠ABC)² = (2/√5)² = 4/5.

sin²∠ABC = (sin∠ABC)² = (1/√5)² = 1/5

Then cos(2∠ABC) = 4/5 - 1/5 = 3/5.

⅗ or 0.6

Step-by-step explanation:

According to the double angle identity:

cos(2X) = 2cos²(X) - 1

Let Angle ABC = X, then

cos(X) = 4/[2sqrt(5)]

cos²(X) = 4²/[2sqrt(5)]²

= 16/20

= 4/5

cos(2X) = 2(4/5) - 1 = 3/5

sorry idk

step-by-step explanation: