Other ways to solve the math.

Option 1: From lecture

(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 720 inches distance to shooter (AB) = 732.3 inches

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

Building is 60 feet away along the horizon line; Bullet hole is 4 feet above the ground. Where is the shooter located? BC2 = 536117 – 518400 distance to shooter = 732.3 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 = 133.1 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)

Option 2:

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 732.32 = 7202 + BC2 BC2 = 732.32 – 7202 BC2 = 536117 – 518400 BC = √17717 (square root) BC = 133.1 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

Option 3:

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– 552.25 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

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 720 inches bc = 4.35 in (previous slide) BC = (720 in * 4.35 in) / 23. 5 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 = 133.3 ft BC = 11.1 ft