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)
