The line of reflection is at 5 on the x-axis. C is at the point,(6,2)
Given:

To find:
The exact value of cos 15°.
Solution:

Using half-angle identity:


Using the trigonometric identity: 

Let us first solve the fraction in the numerator.

Using fraction rule: 

Apply radical rule: ![\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7B%5Cfrac%7Ba%7D%7Bb%7D%7D%3D%5Cfrac%7B%5Csqrt%5Bn%5D%7Ba%7D%7D%7B%5Csqrt%5Bn%5D%7Bb%7D%7D)

Using
:


Answer:
C
Step-by-step explanation:
Answer:
Step-by-step explanation:
In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:
a = -32 ft/s/s
v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)
Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)
t = ?? sec.
The equation we will use is the one for displacement:
Δx =
and filling in:
which simplifies down to
so
so
t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)
Now we use that time in the x-dimension. Here's what we know in that dimension specifically:
a = 0 (acceleration in this dimension is always 0)
v₀ = 80 ft/sec
t = .968 sec
Δx = ?? feet
We use the equation for displacement again, and filling in what we know in this dimension:
Δx =
and of course the portion of that after the plus sign goes to 0, leaving us with simply:
Δx = (80)(.968)
Δx = 77.46 feet
Opposites are the same difference from zero!