- 18°
- The missiles are set up about 200 meters away from the watchtower, facing 18° upwards. How tall is the watchtower
- 200 meters
- 18°
- This a regular trig problem since we have both one angle and one side length. Since we are trying to find the opposite side length, we will using tangent or tan.
- the equation we will be using is Tan(18°)=x/200. First we will cross multiply so that the equation is now 200 times tan(18°) which is 64.9
- So the watchtower is about 65 meters tall
- 200 meters
- 65 meters
- Sir!
- Well done Sergeant, but your not done yet. There one more, it a bit more strenuous.
- 20°
- 50°
- The watchtower is sitting atop a wall that is 130 meters high. At 20° the missiles can hit the wall, and at 50° they can hit the tower. How tall is the tower
- 130 meters
- When we calculate that we get 357.17So the missiles are 357.2 meters away from the cliff. But we are not done yet
- 20°
- 50°
- Okay, the first triangle we have an angle and side length. We have the opposite angle and we are trying to find the adjacent angle. Meaning we using Tan.
- the equation we will be using is Tan(20°)=130/x. First we will cross multiply so that the equation is x=130/tan(20)
- 357.2 meters
- 130 meters
- When we calculate that we get 425.69. the the walls and watchtower combine height is about 426 meters. Now we subtract 130 from 426 which is 296. So the Watchtower is 296 meters tall
- 20°
- 50°
- Now we can do the second triangle. Since we have one angle and an adjacent side length we are using tan.
- the equation we will be using is Tan(50°)=x/357.2. First we will cross multiply so that the equation is x(tan(50) over tan(50) equal to 357.2 over tan(50)
- 357.2 meters
- 130 meters
- 296 meters