Answer:
80000
Step-by-step explanation:
Answer:
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.
To find the probability of the ball landing outside the free throw box but on the court you find the total area of the court and subtract the area of the free throw box.
Then create a ratio with the part of the court not with the free throw box and the total court area. Divide these.
Total area = 25.6 x 15.2 =389.12 square meters
Free throw space = 5.8 x 3.7=
21.46 square meters
389.12-21.46=367.66 square meters
367.66/389.12=0.95
There is a 0.95 probability that it will land on the court outside of the free throw box.