1. Other ways to solve the math.

2. Option 1: From lecture

3. (23.5)2 = (23.9)2 + bc2
Building is 60 feet away along the horizon line; Bullet hole is 4 feet above the ground. Where is the shooter located?
23.9”
Distance to window (Ab) = distance to shooter (AB)
Distance along horizon (Ac) distance to side of building (AC)
23.5”
AC = 60 ft * 12 in/ft = 720 inches
23.9 in = distance to shooter
23.5 in inches
distance to shooter (AB) = inches

4. Now use Pythagorean’s theorem to find BC
732.3
23.9”
23.5”
720 inches
Now use Pythagorean’s theorem to find BC

5. Building is 60 feet away along the horizon line; Bullet hole is 4 feet above the ground. Where is the shooter located?
BC2 = –
distance to shooter = inches
Use Pythagorean’s theorem AB2 = AC2 + BC2
23.9”
AB = distance to shooter
23.5”
AC = distance to building
BC = height of the shooter from the horizon (not from the ground)
AC = 60 ft * 12 in/ft = 720 inches
(732.3 in)2 = (720 in) 2 + BC2
BC2 = (732.3 in)2 – (720 in) 2
BC = √( 536,263 in2 – 518, 400 in2)
(square root)
BC = inches
BC = 11.1 feet
Shooter is 11.1 feet higher than the bullet hole, which is 4 ft. Shooter was 15.1 feet about the ground (on a second floor)

6. Option 2:

7. B is where the shooter is located; find the length of BC
B is where the shooter is located; find the length of BC. The Abc triangle has the same proportions as the ABC triangle
So or
AB = 732.3”
Using Pythagorean’s theorem AB2 = AC2 + BC2
= BC2
BC2 = – 7202
BC2 = –
BC = √17717 (square root)
BC = inches
BC = 11.1 feet
We know that the bullet hole in the seat is four feet above the ground, so the shooter is 15.1 feet above the ground

8. Option 3:

9. Building is 60 feet away along the horizon line; Bullet hole is 4 feet above the ground. Where is the shooter located?
(23.5)2 = (23.9)2 + bc2
Ab2 = Ac2 + bc2
(23.9)2 = (23.5)2 + bc2
bc2 = (23.9)2 - (23.5)2
bc2 = 571.2–
We know two sides of the smaller triangle, so use Pythagorean’s to solve for bc
bc = √18.95 (square root)
bc = 4.35 in

10. Building is 60 feet away along the horizon line; Bullet hole is 4 feet above the ground. Where is the shooter located?
(23.5)2 = (23.9)2 + bc2
Now, set-up a ratio to relate to the larger triangle involving the building.
bc = BC
Distance along horizon distance to side of building
4.35 in = BC
23.5 in inches
bc = 4.35 in (previous slide)
BC = (720 in * 4.35 in) / in
We know that the bullet hole in the seat is four feet above the ground, so the shooter is 15.1 feet above the ground
BC = ft
BC = 11.1 ft